CHAPTER 7 MAXIMA AND MINIMA 7.1 INTRODUCTION The notion of optimizing functions is one of the most important application of calculus used in almost every sphere of life including geometry, business, trade, industries, economics, medicines and even at home. To control the ideas of greatest or least, highest or lowest, farthest or closest, etc., we introduce the technical terms maximum and minimum as applied to functions. However, the terms maximum and minimum without further qualification turn out to be insufficient to cope with all types of optimizing problems. For example, when a stone is thrown upward, it attains a maximum height and a minimum height, and the story ends there; whereas, when a rubber ball is tossed up, it attains a highest point, descends to a lowest point, but then bounces up again to attain another "maximum" height, then falls, bounces up again to still another "maximum" height, and continues this rise and fall process until it comes to rest. These "maxima" in quotes are "local" maxima in the sense that only when restricted to competition in the time interval from bounce to bounce are they guaranteed to be the maximum in the strictest sense of the word. A similar situation shows up in the oscillation of a pendulum which is slowly coming to rest. In this case the bob of the pendulum reaches extreme right and left positions over intervals of time. Of course, the first extreme right position would be the undisputed maximum, while the first extreme left position would be the undisputed minimum; but the extreme positions of each subsequent interval of swing qualify as maxima and minima only relative to the corresponding time interval of swing. In view of the above illustrative situations we are led to introduce the adjectives local and global to distinguish the two type of extreme positions. But the only characteristic which distinguishes the local from the global will be the domain of validity. In fact, the local concept will merely be the global concept if a small part of the given domain is considered. In terms of graph, the global maximum is the y-coordinate of the highest point of the graph. But the local maxima are y-coordinates of the locally highest points of the graph, namely, the points which are highest in competition with only the points of the graph contained within a suitably thin vertical strip centred at the highest point. Similar remarks hold when minimum and lowest replace maximum and highest, respectively. 7. CONCEPT OF LOCAL MAXIMA AND LOCAL MINIMA Let y = f(x) be a function defined at x = a and also in the neighbourhood of the point x = a. Then, f(x) is said to
7. DIFFERENTIAL CALCULUS FOR JEE MAIN AND ADVANCED have a local maximum at x = a, if the value of the function at x = a is greater than the values of the function at all points in the neighbourhood of x = a. Mathematically, f(a) > f(a h) and f(a) > f(a + h) where h > 0 is a small positive arbitrary number. In other words, a function f(x) is said to have a local maximum at x = a if f(a) > f(x) x (a h, a + h) {a}, where h is a small positive arbitrary number. Similarly, f(x) is said to have a local minimum at x = a, if the value of the function at x = a is smaller than the values of the function at all points in the neighbourhood of x = a. Mathematically, f(a) < f(a h) and f(a) < f(a + h) where h > 0 is a small positive arbitrary number. 'turning point' is used both for local maximum or minimum values. (v) A local maximum (minimum) value of a function may not be the greatest (least) value in an interval. (vi) A function can have several local maximum and minimum values, and a local minimum value may even be greater than a local maximum value. Here is a function, defined on the interval [a, b], which has a local maximum at x = x 1 and x = x 3, has a local minimum at x = x and x = x 4, Note that the minimum of the function at x = x 4 is greater than the maximum of the function at x = x 1. At x = b, the value of the function is greater than any local maximum of the function in the interior of the interval under consideration. In other words, a function f(x) is said to have a local minimum at x = a if f(a) < f(x) x (a h, a + h) {a}, where h is a small positive arbitrary number. Note : (i) The above definition is applicable to all functions continuous or discontinuous, differentiable or non-differentiable at x = a. (ii) If the graph of a function f attains a local maximum at the point (a, f(a)), then x = a is called the point of local maximum and f(a) is called the local maximum value. A similar terminology is used for local minimum. (iii) The local maximum and minimum values of a function are also known as relative maxima or relative minima as these are the greatest and least values of the function relative to some neighbourhood of the point in question. (iv) The generic terms for maxima and minima of a function are extremum (plural extrema) or extreme values of the function. The term 'extremum' or (vii) Usefully we deal with functions having only a finite number of extrema in a given finite interval. Consider the function f(x) = x sin 1/x, x 0f(0) = 0. In every interval containing the point 0, it is continuous and has an infinite number of points of extrema. (viii) If a function is strictly increasing or strictly decreasing at an interior point x = a it cannot have an extremum at x = a and vice versa. Test f(x) = {x} for the existence of a local maximum and minimum at x = 1, where {.} represents the fractional part function. Clearly x = 1 is a point of discontinuity. f(1) = 0 Now, f(1 h) > 0 and f(1 + h) > 0, i.e., the value of the function at x = 1 is less than the values of the function at the neighbouring points. Thus, x = 1 is the point of local minimum.
MAXIMA AND MINIMA 7.3 Let f(x) = ì ï x 0 í < x. ï î 1 x = 0 Examine the behaviour of f(x) at x = 0. The graph of y = f(x) is shown below : f(x) has local maxima at x = 0 because f(0) = 1 is greater than every other value assumed by f(x) in the immediate neighbourhood of x = 0. When speaking of a maximum or a minimum at a point x = a, we usually mean a strict extremum. In such a case, there exists a neighbourhood of the point of extremum x = a for all whose points (except the point a itself) there holds the strict inequality f(x) < f(a) for the point of maximum or the strict inequality f(x) > f(a) for the point of minimum If we relax our definition, and the signs > and < in these inequalities are replaced, respectively, by and the point x = a is called a point of non-strict extremum. Consider the graph of a function in the neighbourhood of the point x = a as shown in the figure : We observe that in the given figure, f(a h) f(a) f(a + h). Here, we say that the function has a point of local maximum (non-strict) at x = a. The function shown below has a point of local minimum (non-strict) at x = a, since f(a h) f(a) f(a + h). If f(x) is constant in the neighbourhood of the point x = a then it said to have both a local maximum and a local minimum at x = a. Thus, a function f(x) is said to have a local maximum at x = a if f(x) f(a) x (a h, a + h), when h is a small positive arbitary number. And, it is said to have a local maximum at x = a if f(x) f(a) x (a h, a + h). Extremum at end points A point (a, f(a)) is called an endpoint of the graph of the function f if there exists an interval () containing a such that the domain of f contains every number of the interval (, a) and no number of the interval (a, ), or vice versa. If x = a happens to be an endpoint of the function, then we compare f(a) with appropriate values of the function in either the left or right neighbourhood of x = a. Consider a function f(x) defined in the interval [a, b]. Then, the function is said to have a local maximum at the left end point x = a if f(a) > f(a + h), where h is a small positive arbitrary number. It is said to have a local minimum at the left end point x = a if f(a) < f(a + h). The function is said to have a local maximum at the right end point x = b if f(b h) < f(b). It is said to have a local minimum at the right end point x = b if f(b h) > f(b). However, if a function is not defined at its endpoint then there is no extremum at such a point. We can see from these figures that endpoints (provided the function is defined) are almost always points of local maxima or minima. Consider f(x) = 4 x. Clearly, f( ) = 0 and f() = 0 are local minima of f, since
7.4 DIFFERENTIAL CALCULUS FOR JEE MAIN AND ADVANCED 0 is the least of all the values of the function in the right neighbourhood of x = and the left neighbourhood of x = respectively. x1 a, x1 Let f(x) =. If f(x) x 3,x 1 has a local minima at x = 1, then find all possible values of a. 1xa, x1 f(x) x 3, x1 Since f(1) = 5, the local minimum value of f(x) at x = 1 will be 5. Hence 5 should be smaller than every other value assumed by f(x) in the immediate neighbourhood of x = 1. The function increases above 5 in the right neighbourhood of x = 1. In the left neighbourhood of x = 1, the function is decreasing. So the left hand limit of the function should be either equal to or greater than 5 to have a local minimum at x = 1. f(1 ) f(1) a 5. If a < b < c < d and x R then find the points of extrema of the function, f(x) = x a + x b + x c + x d f (x) = x a + x b + x c + x d é4x -(a + b + c + d) x < a x - a + (b + c + d) a x < b = ê c + d - (a + b) b x < c x - (a + b + c) + d c x < d êë 4x - (a + b + c + d) x ³ d The graph of y = f(x) is shown below : We can observe in the figure that the function has a local minimum or maximum (non-strict) at each x between b and c i.e. b < x < c. It has a local minimum at x = b and c. x = a is not a point of extremum because the function is strictly decreasing at x = a. Similarly the function is strictly increasing at x = d. lim f(x) lim[f(x)] (a R), If x a x a where [ ] denotes greatest integer function and f(x) is a non constant continuous function, then show that f(x) has a local minimum at x = a. Clearly lim[f(x)] is an integer and LHL xa and RHL should be same for existence of lim f(x). x a The function should approach an integer from both sides of x = a. Let lim[f(x)] = n (n I) then lim f(x) = n. xa xa Clearly the function should be just greater than n on both sides of x = a since f(x) is continuous at x = a. f(x) has local minimum at x = a. If f(x) = (sin x 1) n ( + cos x), then prove that x = / is a point of maximum, if n is an odd positive integer. We have f( / ) = 0. If x = a is a point of local maximum, then f(a h) < 0 and f(a + h) < 0 f(/ h) < 0 andf(/ + h) < 0 f(/ h) = ( ve) n... (1) f(/ + h) = ( ve) n... () f(/ h) and f(/ + h) are negative when n is odd. So, f(x) has local maximum at x = / when n is odd. Note that both f(/ h)and f(/ + h)are positive if n is even in which case the function has a local minimum at x = /. So, f(x) has maxima or minima at x = / according as n is odd or even. 7.3 FERMAT THEOREM It appears that at the maximum and minimum points the tangent lines are horizontal and therefore each has slope 0. The following theorem says that this is always true for differentiable functions.
MAXIMA AND MINIMA 7.5 Fermat Theorem. If a belongs to an open interval in the domain of f, if f(a) exists, and if (a, f(a)) is a point of local extremum (either a maximum or a minimum), then f(a) = 0. Geometrically, Fermat theorem says that if there is a tangent at a highest point or lowest point of the graph and further if the point of contact is not an end point of the graph, then the tangent line is necessarily horizontal. Proof : Geometrically this theorem is obvious. We shall prove it, only in the case that (a, f(a)) is a local maximum point, since a similar proof (with the inequalities reversed) is valid for a local minimum point. Since (a, f(a)) is a local maximum point, f(a h) f(a), and f(a + h) f(a) for all positive h in some open interval containing 0. Thus f(a h) f(a) 0. Since h is positive, f(a h) f(a) 0 and h f(a h) f(a) lim 0 h0 h and, f(a h) f(a) 0 and h f(a h) f(a) lim 0 h0 h Since f(a) exists, the two limits above must have a common value which is f(a). Thus f(a) is both greater than or equal to zero and also less than or equal to zero. The only number which satisfies both of these conditions is zero, hence f(a) = 0. This completes the proof. This theorem is instrumental in discovering both local and global extrema. Suppose x 3 + ax + bx + c satisfies 50 f ( ) = 10 and takes the extreme value where x =. 7 3 Find the value of a, b and c. f(x) = x 3 + ax + bx + c f ( ) = 8 + 4a b + c = 10 4a b + c + = 0...(1) f ' (x) = 3x + ax + b f '(/3) = 0 by Fermat theorem 4 4 4 a 3 + a + b = 0 + + b = 0 9 3 3 3 4 + 4a + 3b = 0...() 50 Also f(/3) = 7 8 4 a 50 This gives + + + c = 7 9 3b 7 8 + 1a + 18b + 7c = 50 1a + 18b + 7c = 4 4a + 6b + 9c = 14...(3) Solving (1), () and (3) we get a = 1, b = 0 and c =. Given that the function f(x) = (x p) + (x q) + (x r) has a minimum, find the corresponding value of x. We have, f(x) = (x p) + (x q) + (x r)...(1) It is a differentiable function for x ( ). Hence by Fermat theorem, a minimum can be attained when f (x) = 0. Differentiating (1) w.r.t. x, we get f(x) = (x p) + (x q) + (x r)...() f(x) = 0 when (x p) + (x q) + (x r) = 0 3x (p + q + r) = 0 x = 3 1 (p + q + r). Since a single value of x is obtained, without further investigation we can say the minimum is attained at x = 3 1 (p + q + r). Check whether the function f(x) = 1 x, x [ 1, 0] satisfies the conditions of Fermat theorem. The given function is strictly decreasing in [ 1, 0] and consequently attains maximum at x = 1 and minima at x = 0. Since, these points are not interior points of the interval [ 1, 0] we can not apply Fermat theorem to assert that f( 1) = f '(0) = 0 In fact f '( 1) = 4 and f '(0) = 0. We see that f '(0) = 0 but this inference could not have been drawn using Fermat theorem (which is not applicable in this problem). If f'(x) = 0 at x = a, then we say that x = a is a stationary point of the function f. Also, f(a) is then said to be a stationary value of f. We say so because the rate of change of the function is zero at such a point. If f(x) = x 3 + ax + bx + c has stationary point at x = 1 and x = 3. Find a, b, c.
7.6 DIFFERENTIAL CALCULUS FOR JEE MAIN AND ADVANCED Since f(x) has stationary point at x = 1 and x = 3, f( 1) = 0 = f(3) f(x) = 3x + ax + b f( 1) = 3 a + b = 0 f(3) = 7 + 6a + b = 0 a = 3, b = 9, c R. Show that the function f defined by f(x) = x 4 x 3 + 3x 1 has a stationary point in the interval ( 1, ). We have f (x) = 4x 3 6x + 3 and f ( 1) = 7, f () = 11. Since f is continuous in [, 3], f has a stationary point a in ( 1, ). Evidently, f(a) = 0. Thus, a is root of the equation 4x 3 6x + 3 = 0. We find the set of values of x for which f (x) = 0 and carry out investigation to locate the points of extrema. However, vanishing of f' (c) is only a necessary condition and not a sufficient condition for x = c to be a point of extremum. Consider the graph of y = x 3. We can see that f'(0) = (3x ) x = 0 = 0 but x = 0 is not a point of extremum. Indeed, no matter how close the point x is to origin, we will always have x 3 < 0 when x < 0 and x 3 > 0 when x > 0. The graph is rising for x < 0 and also for x > 0, as shown in the figure. Critical points Critical points are interior points in the domain of the function where either f(x) = 0 or f(x) does not exist. In Fermat theorem we have considered the critical points at which the derivative is equal to zero i.e stationary points. Let us now deal with the critical points at which the derivative does not exist. Find the critical points of the function f(x) = 4x 3 6x 4x + 9 if (i) x [0, 3] (ii) x [ 3, 3] (iii) x [ 1, ]. f(x) = 1(x x ) = 1(x ) (x + 1) f(x) = 0 x = 1 or (i) If x [0, 3], x = is the critical point. (ii) If x [ 3, 3], then we have two critical points x = 1,. (iii) If x [ 1, ], then there is no critical point as both x = 1 and x = become endpoints. Critical points are always interior points of an interval. Find the number of critical points for f(x) = max (sin x, cos x), x (0, ). The graph of y = f(x) is shown below : Thus, Fermat theorem tells us that a local extremum of a differentiable function f can occur only at a stationary point, but it does not say that a local extremum must occur at each stationary point. The converse of Fermat's theorem is false in general. Furthermore, there may be an extreme value at x = a even when f '(a) does not exist. Theorem Suppose that f is a function defined on an interval containing the number x = a. If the function has a point of local extremum at x = a, then either f(a) = 0 or f(a) does not exist or a is an endpoint of the interval. We can see from the figure that f(x) = 0 at x = p and f(x) does not exist at x = p 4, 5p 4. Thus, the critical points of f(x) are x = p 4, p, 5p 4. Hence, f(x) has three critical points. Find the critical points of the function f(x) = x x and find whether these points are points of extrema. We have f(x) = x x ì ïx - x x 0,x³ f(x) = í ïî x- x 0< x<
MAXIMA AND MINIMA 7.7 ìï (x - 1) x < 0,x > f(x) = í ïî (1- x ) 0< x < f(x) = 0 at x = 1 and f(x) does not exist at x = 0,. Thus, the critical points of f(x) are x = 0, 1,. Note that x = 1 is also a stationary point. The graph of y = f(x) is shown below : 1. In each of following cases, identify if x = a is point of local maxima, minima or neither of them. (i) (ii) (iii) (iv). Draw the graph of the following functions and locate the points of extrema : (i) y = x 1 1 (ii) y = x 1 (iii) y = x + 1 + x 3. Find the points of extrema of the function f(x) = x + 5 x 3. 7.4 THE FIRST DERIVATIVE TEST From what has been discussed in the previous section, From the graph, we can see that x = 1 is a point of local maximum and x = 0, are points of local minima. It is also quite possible for a continuous function f to have no local extremum at a point a where f'(x) does not exist. For example, the function y = 3 x has no derivative at x = 0. (y' as x 0). Thus, x = 0 is a critical point of the function where the derivative does not exist. At this point the function has neither a maximum nor a minimum because f(0) = 0, f(x) < 0 for x < 0, f(x) > 0 for x > 0. We see that f(x) is strictly increasing at x = 0 and hence, has no extremum at x = 0. A ìx+1, x< ï 4. Let f(x) = ía, x =, Find a if f(x) has local ï ïî a-x, x> maximum at x =. 5. Find the points of extrema for the function y = x + x a. ì ï(x + l) x < 0 6. If f(x) = í, find possible values ï î cosx x ³ 0 of such that f(x) has local maxima at x = 0. 7. Show that f(x) = 1 (x 1) 1/3 has a critical point at x = 1, but the function has neither a maximum nor a minimum at x = 1. 8. The greatest integer function f(x) = [x], defined for all values of x, assumes a local maximum value of 0 at each point of [0, 1). Could any of these local maximum values also be local minimum values of f? Give reasons for your answer. 9. If a function f is continuous in an interval I and if f(a) > 0, f(b) < 0 for some a, b in I, then prove that f has a critical point in I. 10. Suppose f is differentiable on [a, b], and f(a) < k < f(b) or f(a) > k > f(b). Prove there exists c (a, b) such that f(c) = k. it follows that not for every critical point does a function have a maximum or a minimum. However, if at some interior point the function attains a maximum or a minimum, this point is definitely critical. And so to find
7.8 DIFFERENTIAL CALCULUS FOR JEE MAIN AND ADVANCED the extrema of a function, do as follows: find all the critical points, and then, investigating separately each critical point, find out whether the function will have a maximum or a minimum at that point, or whether there will be neither maximum nor minimum. Usually the behaviour of a continuous function y = f(x) at a critical point x = a can be determined from the algebraic sign of the derivative near a. Consider the following graphs in the neighbourhood of a critical point x = a : (i) If f changes from positive to negative at a (f > 0 for x < a and f < 0 for x > a), then f has a local maximum value at a. (ii) If f changes from negative to positive at a (f < 0 for x < a and f > 0 for x > a), then f has a local minimum value at a. (iii) If f does not change sign at c (f has the same sign on both sides of c), then f has no local extreme value at c. The First Derivative Test Suppose that x = a is a critical point of a continuous function y = f(x). (i) If f(x) changes sign from positive to negative at x = a, then f has a local maximum at x = a. (ii) If f(x) changes sign from negative to positive at x = a, then f has a local minimum at x = a. (iii) If f(x) does not change sign at x = a (that is, if f(x) is positive on both sides of x = a or negative on both sides), then f has no local maximum or minimum at x = a. In part (i), when the sign of f(x) changes from positive to negative at x = a, f is increasing on the left of x = a and decreasing on the right of x = a, it follows that f has local maximum at x = a. It is easy to remember the first derivative test by visualizing the above diagrams. Informally, the derivative test says, "If the derivative changes sign at x = a, then the function has either a local minimum or a local maximum." To decide which it is, just make a crude sketch of the graph near (a, f(a)) to show on which side of x = a the function is increasing and on which side it is decreasing. Mathematically, if x = a is a critical point of the function f(x) and the inequalities f(a h) > 0, f(a + h) < 0 are satisfied, where h is a small positive arbitrary number, then the function f(x) possesses a maximum at the point x = a.. If f(a h) < 0, f(a + h) > 0, then the function f(x) possesses a minimum at the point x = a. If the signs of f(a h) and f(a + h) are the same, then the function f(x) does not possess an extremum at the point x = a. Saddle point Let x = a be a stationary point of the function y = f(x) i.e. f(a) = 0. If f(x) has the same sign on either side of x = a, then f has a saddle point at x = a. The point (a, f(a)) is known as a saddle point, as it can be envisioned as a level resting spot on the side of an ascending or descending portion of a functional curve. A saddle point is a special case of point of inflection. For example, if f(x) = tan 3 x, then f'(x) = 3tan x sec x and f'(0) = 0, f(0 h) > 0, f(0 + h) > 0.
MAXIMA AND MINIMA 7.9 Hence, the graph is rising for p < x < 0 and also for 0 < x < p. The origin is a saddle point of f. The function has a point of inflection at x = 0. There is no extremum at x = 0 on the graph of f. Let f(x) = ln (x ) x + 4x + 1. Investigate the function for extremum at x = 3. The function f(x) = ln (x ) x + 4x + 1 is defined for x >. Let us find its derivative : f(x) = x x + 4 = (x 1)(3 x). x The point x = 3 is a critical point of the given function. Sign scheme of f '(x) We can see from the figure that f(0) = 1, and f(x) < 1 for all x different from zero. Let f(x) = 3x x The given function has no derivative at the point x = 0. Since for all x < 0, f(x) < f(0), and for all x > 0, f(x) > f(0), the given function has no extremum at the point x = 0. On the interval (, 3), the derivative f(x) is positive and, consequently, on the interval the function increases. And on the infinite interval (3, ), the derivative f(x) is negative and, therefore, the function f(x) decreases on that interval. The point x = 3 is a point of local maximum. By considering the function f defined as 1 f(x) = x 4 sin, if x 0, f(0) = 0, x it can be seen that the conditions of the first derivative test are sufficient but not necessary. In fact, f is derivable everywhere, f '(x) does not change sign from negative to positive as x passes through 0 and yet f has a minimum at x = 0. We have considered the critical points at which the derivative is equal to zero i.e. the stationary points. Let us now deal with the critical points at which the derivative does not exist. 3/ The function f(x) = 1 x 3 has no derivative at 1 3 3 x = 0, since f '(x) = 1x x approaches infinity at x = 0, but the function has a maximum at this point. 1/ We cannot say that the change of sign of the derivative always helps in determining maxima/minima. For example, let x, x 0 f(x) = 5, x 0 sinx,x 0 f'(x) = x, x 0 cosx, x 0 f'(x) does not exist and f is discontinuous. Here also the derivative is changing sign from negative to positive, but x = 0 is not a point of minimum. On the contrary, it is a point of maximum since f(0) = 5 and f(0 h) < f(0) and f(0 + h) > f(0). This type of difficulty appears in case of discontinuous functions. Test for Local Maximum/Minimum when f(x) is not differentiable at x = a Case 1: When f(x) is continuous at x = a and f'(a ) and f'(a + ) exist and are non-zero, then f(x) has a local maximum or minimum at x = a according to the following : If f'(a ) > 0 and f'(a + ) < 0, then x = a is a point of local maximum. If f'(a ) < 0 and f'(a + ) > 0, then x = a is a point of local minimum. Case : When f(x) is continuous and one or both of f'(a ) and f'(a + ) are either zero or does not exist, then
7.10 DIFFERENTIAL CALCULUS FOR JEE MAIN AND ADVANCED we should consider the signs of f'(a h) and f'(a + h), where h is a small positive arbitrary number. If f'(a h) > 0 and f'(a + h) < 0, then x = a is a point of local maximum. If f'(a h) < 0 and f'(a + h) > 0, then x = a is a point of local minimum. Case 3: If f(x) is discontinuous at x = a, then we should find about the existence of local maxima/minima using the basic definition of local maxima/minima i.e. compare the values of f(x) at the neighbouring points of x = a. We should find the values of f(a ), f(a + ) and f(a). If f(a) is the greatest of all these values, then there is a local maximum at a. If f(a) is the least of all these values, then there is a local minimum at a. It is also advisable to draw the graph of the function in the vicinity of the point x = a, because the graph would give us a clear picture about the existence of local maxima/minima at x = a. If f(a) is undefined, then there is no question of extremum at x = a. Test the function f(x) = (x ) /3 (x + 1) for extrema at x = 1 and. 10 x 1 We find f '(x) =. 3 3 x. The critical points are x = 1 (the derivative is zero) and x = (the derivative does not exist). The function is continuous. The inequalities f '(1 h) > 0, f '(1 + h) < 0, f '( h) < 0, f '( + h) > 0 hold at a sufficiently small h > 0. Consequently, at the point x = 1 the function possesses a local maximum and at the point x = it possesses a local minimum since f( h) < 0 and f( + h) > 0. Let f(x) = ì 3 ï x + x + 10x, x < 0 í ï- î 3sinx, x ³ 0. Investigate the function for extremum at x = 0. Clearly f(x) is continuous at x = 0 but non- differentiable at x = 0. We have 3 f'(0 f( h) f(0) h h 10h0 )= lim = lim 10 h 0 h h0 h f( h) f(0) -3sinh But f'(0 + ) = lim = lim =-3 h 0 h h 0 h Since f'(0 ) > 0 and f'(0 + ) < 0, x = 0 is the point of local maximum. Alternative : f(x) is continuous at x = 0. ì ï3x + x - 10 x < 0 f(x) = í ï î3cosx x > 0 f(0 ) = 10 and f(0 ) = 3. Thus, f(x) is non-differentiable at x = 0 x = 0 is a critical point. Also the derivative changes sign from negative to positive. So x = 0 is a point of local minima. Let f(x) = ì ï x, x 0 í. Investigate the ï îsinx, x > 0 function for extremum at x = 0. We sketch the graph of f(x) in the vicinity of x = 0. We see that x = 0 is a point of local minimum. Alternative : f'(x) = { x, x < 0 cosx, x> 0 f'(0) does not exist but f is continuous at x = 0. x = 0 is a critical point where f '(0 ) = 0 and f '(0 + ) = > 0. On the left of x = 0, f '(0 h) < 0. Thus, the derivative changes sign from negative to positive. Hence, x = 0 is a point of minimum. Test the function f(x) = 1 (x ) 4/5 for extremum at x =. We find f(x) = 4 5 (x 4 ) 1/5 = 5 5 x. The derivative does not exist at x = (critical point). The one-sided derivatives at x = do not exist. But f ( h) > 0 and f ( + h) < 0, where h is a small positive arbitrary number. Hence, we infer that the function possesses a local maximum at x =. Investigate the following functions for extrema :
MAXIMA AND MINIMA 7.11 (i) (ii) x, x 0, f(x) 3x 5, x 0 x 3, x 0, f(x) 4, x 0, x 0 (i) Though the derivative f'(x) 3, x 0 exists at all points, except the point x = 0, and changes sign from minus to plus passing through the point x = 0, there is no minimum here because f(0 h) < f(0) = 5. Figure (a) Figure (b) This is explained by the fact that the function is discontinuous at the point x = 0. (b) Here the derivative f'(x) = 4x(x 0) also exists at all points except at x = 0, and it changes sign from minus to plus when passing through the point x = 0. Nevertheless, we have here a maximum but not a minimum, which can readily be checked. It is explained by the fact that the function is discontinuous at the point x = 0. x cos,x 0 Let f(x) =. Find the xa, x0 values of a if x = 0 is a point of local maximum. of a. We draw the graph of with different values Clearly, f(x) increases before x = 0 and decreases after x = 0. f(0) = a. Note that in figure (a) for a = 0, x = 0 is the not point of maximum. The graph of y = x + a must shift atleast 1 unit upwards, for x = 0 to be the point of maximum. For x = 0 to be the point of local maxima, we have f(0) lim f(x) x 0 æpx ö f(0) lim cos + ç a 1 x 0 è ø Hence, a 1. Let ì ï x + a - 9a 9,if x< f(x) = í ï î x 3, if x ³ Find the value of 'a' for which f(x) has local minimum at x =. We have, ì ï x + a - 9a 9,if x< f(x) = í îï x 3, if x ³ Given f(x) has local minima at x =. Since, f(x) = x 3 is strictly increasing for x lim f (x) ³ f () - x lim h0 f( h) { f() = 3 = 1} lim h 0 { h + a 9a 9} 1 a 9a 10 0 (a + 1) (a 10) 0 a 1 or a 10. Given a continuous function sin x at x 0, f(x) 1 x at x 0. 0 Show that f(x) has a minimum at the point x = 0, but is not monotonic either on the left or on the right of x = 0. For x 0, f(x) > 0, and f(0) = 0, hence f(0) is a minimum. For x > 0 the derivative f'(x) = sin x 1 + x 1 cos x 1 is positive at the points 1 x = and negative at the p n points 1 x =. Hence, f is not monotonic on (n + 1) p the right of x = 0. The case x < 0 is investigated analogously.
7.1 DIFFERENTIAL CALCULUS FOR JEE MAIN AND ADVANCED Extremum at Endpoints Let y = f(x) be continuous at the endpoint of the interval [a, b] At a left endpoint a : If f < 0 (f > 0) for x > a, then f has a local maximum (minimum) value at a. the sign of f'(x) remains unchanged as x passes through the point under consideration, then the function f(x) has no extremum at this point. Find the local minimum and maximum values of the function f(x) = 3x 4 4x 3 1x + 5. f(x) = 1x (x ) (x + 1) f(x) = 0 x = 1, 0,. The critical points are : x 1 = 1, 0 and x =. Sign scheme of f(x) At right endpoint b : If f < 0 (f > 0) for x < b, then f has a local minimum (maximum) value at b. Find the extrema of the function f(x) = 3x (x 1) 3/ at x = 1. Since (x 1) 3/ is real only if x 1, the domain of f is the set [1, ). Thus x = 1 is the left endpoint of the graph of f. Note that f is continuous a x = 1. 3 We havef(x) = 3 (x 1) 1/. Since f (1 + h) > 0, x = 1 is a point of local minimum. Steps for finding the extrema of a function defined in an interval Let f(x) be defined and continuous on a certain interval (a, b) and have a derivative everywhere in the interval (a, b) except possibly at a finite number of points, and have a finite number of stationary points. Then, in order to find the extrema of the function, it is necessary: (i) To find the critical points of the function f(x), i.e. the points at which either f'(x) = 0 or f'(x) does not exist; (ii) To investigate the sign of the derivative f'(x) in a certain neighbourhood of each critical point. And, if, as x passes through such a point, f'(x) changes sign, then f(x) has an extremum at this point. If minus is changed to plus, then the function has a local minimum at this point; if the sign is altered from plus to minus, then a local maximum is the case. But if f(x) changes from negative to positive at 1, so f( 1) = 0 is a local minimum value by the first derivative test. Similarly, f changes from negative to positive at, so f() = 7 is also a local minimum value. f(0) = 5 is a local maximum value because f(x) changes from positive to negative at 0. In case of continuous functions, points of maxima and minima are found alternately. Show that f(x) = x 3 6x + 1x 8 does not have any point of local maxima or minima. f(x) = x 3 6x + 1x 8 f(x) = 3(x 4x + 4) f(x) = 3(x ) f(x) = 0 x = Clearly f(x) does not change sign about x =. f( + ) > 0 and f( ) > 0. So f(x) has no point of maxima or minima. In fact, f(x) is a strictly increasing function for x R and x = is a saddle point or point of inflection of the function. It will be clear from this example that neither maxima nor minima values can arise from the vanishing of such factors of f(x) that have even indices. Find the extremum of the function f(x) = x 3 (x 3), x R. We find the derivative of the given function:
MAXIMA AND MINIMA 7.13 5x 6 f'(x) = 3 and determine the critical points. 3 x f'(x) = 0 if x = 6 and f'(x) does not exist at x = 0. 5 Thus, the critical points are : x 1 = 0 and x = 6 5. Note that f is continuous everywhere. We now investigate the sign of the derivative f'(x) in a certain neighbourhood of each critical point and tabulate the results : Sign of f'(x) + + 0 6 5 x f'(x) f(x) < x < 0 + increases x = 0 does not exist local maximum 0 < x < 6 5 decreases x = 6 5 0 local minimum 6 5 < x < + increases Find the local maximum and minimum values of the function g(x) = x + sin x, 0 x To find the critical point of g, we differentiate: g(x) = 1 + cos x 1 So g(x) = 0 when cos x =. The solutions of this equation are /3 and 4/3. Because g is differentiable everywhere, the only critical points are /3 and 4/3 and so we analyze g in the following table. Interval g(x) = 1 + cos x g (x) 0 < x < /3 + increasing on (0, /3) /3 < x < 4/3 decreasing on (/3, 4/3) 4/3 < x < + increasing on (4/3, ) Because g(x) changes from positive to negative at 3the first derivative test tells us that there is a local maximum at /3 and the local maximum value is g(/3) = p p + sin 3 3 = p æ 3ö + p 3 3 ç = + 3.83 è ø 3 Likewise, g(x) changes from negative to positive at 4/3 and so g(4/3) = 4 p 4p + sin 3 3 = 4 p æ 3ö 4 + p 3 3 ç - = -.46. è ø 3 The specification of the change of sign of the derivative as x passes through the boundaries of the intervals of monotonicity, that is through the points x i, indicates which of these points are points of maxima and which points of minima. It may also turn out that some of the points x i are not points of extremum. This is the case when the function has the same character of monotonicity in two adjoining subintervals (x i 1, x i ) and (x i, x i+1 ) separated by the point x i, i.e. when the derivative has the same sign in them. In this case x i is not a point of extremum of the function (for instance, for the function y = x 3, the point x = 0 belongs to this type). The substitution of the critical numbers x = x i into f(x) yields the corresponding values of the function : f(x 1 ), f(x ),..., f(x n ) As has been already mentioned, not all the them are necessarily extreme. If the values f(x 1 ), f(x ),...f(x n ) have been computed and the endpoint values f(a) and f(b) have also been found, the character of variation of the function becomes clear without investigating the sign of the derivative. For, since we know that the function has no extrema within each of the subintervals (a, x 1 ), (x 1, x ),..., (x n, b) and thus is monotonous on them, the comparison of the values of the function at the end points of each subinterval indicates directly whether the function increases or decreases on that subinterval. It depends on the circumstances which of the techniques should be used. It is sometimes more convenient to consider the sign of the derivative in the subintervals, but sensually we are interested in the values of the function at the critical points the second approach may prove preferable.
7.14 DIFFERENTIAL CALCULUS FOR JEE MAIN AND ADVANCED 3 Find the extrema of the function f(x) = 3 x x. The function is defined and continuous for all x. Now, f'(x) 1 x 3. x From the equation f '(x) = 0 we find the roots of the derivative : x = 1. Furthermore, the derivative goes to infinity at the point x = 0. Thus, the critical points are x 1 = 1, x = 0, x 3 = 1. The substitution of the critical points into f(x) yields the corresponding values of the function : f( 1) =, f(0) = 0, f(1) =. The endpoint values are f( f( Thus, f increases from f( to f( 1) = and then decreases from f( 1) = to f(0) = 0. Hence, the function has a maximum at x = 1. Similarly, f decreases from f( 1) to f(0) = 0 and then increases from f(0) = 0 to f(1) = 1. Hence, the function has a local minimum at x = 0. Further investigation shows that the function has a local maximum at x = 1. If the function f(x) = ax e bx has a local maximum at the point (, 10) then find a and b. f() = 10, hence ae b = 10. ae b = 5...(1) f ' (x) = a [e bx bx e bx ] = 0 f ' () = 0 a(e b be b ) = 0 ae b (1 b) = 0 b = 1/ From (1) if b = 1/, a = 5e or a = 0 (rejected) a = 5e and b = 1/. Find the extremum of the function f(x) = x x We find the derivative f(x) = 1 4x 1...(1) x x Then we equate the derivative f(x) to zero : 1 4x 1 = 0 x x From this we find a critical point : x = 1/4. The denominator of expression (1) is positive for all x R. Consequently, the function has no other critical point except for x = 1/4. It can be seen from expression (1) that f(x) > 0 for x > 1/4 and f(x) < 0 for x < 1/4 i.e., when the derivative passes through the point x = 1/4 it changes sign from minus to plus. Consequently, x = 1/4 is a point of local minimum, with f(1/4) = 15 / 8. Find the points of extrema of the function f(x) = + x 4, 1 x 1 = 4 x, 1 < x = x 5, < x < 3 = 0, x = 3. Also sketch the curve y = f(x). The function is continuous at all points except x =. f'(x) = 4x 3, 1 < x < 1 = x, 1 < x < =, < x < 3. Thus, the critical points of f(x) are x = 0 (points where f'(x) vanishes) and x = 1,, 3 (points where f'(x) does not exist) Now, we have f'(0 ) < 0 and f'(0 + ) > 0 local minima at x = 0 f'(1 ) > 0 and f'(1 + ) < 0 and f(1 ) = f(1) = 3, f(1 + ) = 3 local minima at x = 1. Since f is discontinuous at x =, we use the basic definition test rather then the first derivative test. and f( ) = f() = 0, f( h) > 0, f( + ) = 3 no extremum at x =. At the end points, we have f( 1) > f( 1 + ) local maxima at x = 1 and f(3) < f(3 ) local maxima at x = 3. The above results can be easily understood from the graph shown above.
MAXIMA AND MINIMA 7.15 Let f(x) = ax b + c x, where a > 0, b > 0, c > 0. Find the condition if f(x) attains the minimum value only at one point. b (a c)x, x 0 b f(x) = b (c a)x, 0 x a b (a c)x b, x a 1. Prove that the function x 3 3x 36x + 10 has a maximum when x =, and a minimum when x = 3.. Find the local maximum and minimum value of f(x) = (x 1) (x ) (x 3). 3. Find the local maximum and minimum value of f(x) = x 3 1x + 36x 0. 4. The graph of the derivative f ' of a function f is shown. (i) On what intervals is f increasing or decreasing? (ii) At what values of x does f have a local maximum or minimum? 5. Prove that the function 4x 3 18x + 7x 7 has no maxima or minima. 6. Show that 5 is a critical point of the function g(x) = + (x 5) 3 but g does not have a local extremum at 5. 7. Find the points of local maxima or minima of the following functions : 13. Find the extrema of the given functions : (i) y = x 3 3x (ii) y = x x (iii) y = x ln (1 + x ) 14. Find the points of extrema of the followings functions: The figures clearly indicate that for exactly one point of minima, we must have a c. B (i) f(x) = (x 1) 3 (x + ) (ii) f(x) = sin x x (iii) f(x) = x 3 + x + x + 1. 8. For a certain curve dy = (x 1) (x ) (x 3) dx 3 (x 4) 4. Discuss the character of the curve at the points x = 1, x =, x = 3, x = 4. 9. Prove that if at the point of a minimum there exists a right hand derivative, then it is nonnegative, and if there exists a left hand derivative, then it is non-positive. 10. Determine the points of extremum of the function y = 3x 1 x 3. 11. A differentiable function f has only one critical point x = 5. Identify the relative extrema of f at the critical point if f (4) =.5 and f (6) = 3. ì ï1/x (x > 0), 1. Show that the function y = í ï î 3x (x 0) has a minimum at the point x = 0, though its first derivative does not change sign when passing through this point. (i) f(x) = 3 x + + x + 5 x x 3 x 0 4 x 0 (ii) f(x) = x 3 0 x 1 4x x1 A
7.16 DIFFERENTIAL CALCULUS FOR JEE MAIN AND ADVANCED 15. The graph of the first derivative f' of a function f is shown. (i) On what intervals is f increasing? (ii) At what values of x does f have a local maximum or minimum? (iii) On what intervals is f concave up? (v) What are the x-coordinates of the inflection points of f? 16. Prove that the function f(x) = x 101 + x 51 + x + 1 does not have an extremum. (x 1) 17. Prove that the function 3 has a maximum (x 1) value, and a minimum value 0. 7 18. Show that the function f, defined by f(x) = x 5 5x 4 + 5x 3 1, has a maximum value when x = 1, a minimum value when x = 3 and neither when x = 0. 19. Find the points of maximum and minimum of the curve for which dy = (x ) 3 (x 3) dx 4 (3x 4) 5 (4x 5) 6. 0. Let f(x) = sinx (1 + cosx), x (0, ). Find the number of critical points of f(x). Also identify which of these critical points are points of maxima/ minima. 7.5 THE FIRST DERIVATIVE PROCEDURE FOR SKETCHING THE GRAPH OF A CONTINUOUS FUNCTION Step 1. Compute the derivative f'(x) and determine the critical points of f, that is, where f'(x) = 0 or f '(x) does not exist. Step. Substitute each critical point into f(x) to find 1. Discuss the extremum of 1sinx, x0 f(x) = at x = 0. x x1, x0. Prove that the function x sin (1/ x) for x 0, f(x) 0 for x 0. has a minimum at the point x = 0 but not a strict minimum. 3. A certain function y = f(x) has the property that y' = sin y + y + x. Show that at a critical point the function has a local minimum. 4. If dy n p (x a) (x b) 1, where n and p are dx positive integers, show that x = a gives neither maximum nor minimum values of y, but that x = b gives a minimum. 5. Suppose f is a differentiable function with (x 1) n(x 3x ) f'(x) =. (x ) Find all critical numbers of f and determine whether each corresponds to a local maximum, a local minimum or neither. 6. Find constants A, B, C and D that guarantee that the graph of f(x) = 3x 4 + Ax 3 + Bx + Cx + D will have horizontal tangents at (, 3) and (0, 7). There is a third point that has a horizontal tangent. Find this point. Then, for all three points, determine whether each corresponding to a relative maximum, a relative minimum or neither. 1/3 7. Let f(x) = x Ax B,where A and B are positive constants. Find all critical numbers of f and classify the corresponding value of f as a relative maximum, a relative minimum, or neither. the y-coordinate. Plot these critical points on the coordinate plane. Step 3. Determine where the function is increasing or decreasing by checking the sign of the derivative on the intervals whose endpoints are the critical points found in step. Step 4. Sketch the graph so that it rises on the intervals where f'(x) > 0, passes through the critical points, and has a horizontal tangent where f'(x) = 0 Sketch the graph of f(x) = x 3 + 3x 1x 5.
MAXIMA AND MINIMA 7.17 We begin by computing and factoring the derivative : f(x) = 6x + 6x 1 = 6(x + )(x 1) We see that f '(x) exists for all x, and from the factored form of the derivative we see that f '(x) = 0 when x = and when x = 1. The corresponding y-coordinates are found as follows : f( ) = 15; critical point (, 15) f(1) = 1; critical point (1, 1) Next, to find the intervals of increase and decrease of the function, we plot the critical points on a number line and check the sign of the derivative at values to the left and right of and 1. We find that f is increasing and decreasing, as indicated in the figure. + + 1 The arrow pattern in figure suggests that the graph of f has a relative maximum at (, 15) and a relative minimum at (1, 1). We begin the sketch by plotting these points on a coordinate plane. We put a "cap" at the relative maximum (, 15) and a "cup" at the relative minimum (1, 1), as shown in figure (a). Finally, we note that because f(0) = 5, the graph has its y-intercept at (0, 5). We complete the sketch by drawing the curves so that it increases for x < to the relative maximum at (, 15), decreases for < x < 1, passing through the y-intercept (0, 5) on its way to the relative minimum at (1, 1), and then increases again for x > 1. The completed graph is shown in figure (b). Determine where the function f(x) = (nx)/x is increasing and where it is decreasing. Sketch the graph of f. f is defined only for x > 0. x(1/ x) ( nx).1 1nx f'(x) x x Because x > 0 for all x > 0, it follows that f'(x) = 0 only when n x= 1; thus e is the only critical point of f. We use the first derivative test and look at the sign of f '(x) to the left of e (at x = 1, for example) and to the right (at x = 3, for example). We see that f'(x) > 0 for x < e and f '(x) < 0 for x > e. Thus, f is increasing to the left of e and decreasing to the right, as shown in the figure. We now have most of the information we need to sketch the graph of f. In particular, we know that the graph lies entirely to the right of the y-axis (because f is defined only for x > 0) and that it rises to a relative maximum at x = e, after which it falls indefinitely. nx Moreover, because we obtain x 0 only for x = 1, it follows that the graph crosses the x-axis on the way up but not on the way down, and this causes us to suspect that the graph must "flatten out" in some way as x But what happens at the two "ends" of the graph, as x 0 + and as x? As x approaches 0 from the right, n x decreases without bound and f(x) = ( n x)/x does the same, so the graph approaches the negative y-axis asymptotically. As x increases without bound, we have seen that f(x) decreases, but by how much? The graph cannot cross the x-axis a second time because n x = 0 only for x = 1, so the graph must "flatten out" in some way. Using limits we find that f(x) as x Sketch the graph of f(x) = cos x + cosx on [0, ] The derivative of f is f'(x) = cos x sin x sin x Because f'(x) exists for all x, we solve f '(x) = 0 to find
7.18 DIFFERENTIAL CALCULUS FOR JEE MAIN AND ADVANCED the critical points : cos x sin x sin x = 0 sin x(cos x + 1) = 0 sin x = 0 or cos x = 1 The critical points on the interior of the interval [0, ] are x = 4, and. 3 3 æp ö Because f ç = 1 æ4p ö 1, f()=0, and f ç =, è 3 ø 4 è 3 ø 4 The points corresponding to these are æp 1ö æ4 ç, -, (, 0) and p 1ö ç,- è 3 4 ø è 3 4ø Next, we plot the critical points on a number line and determine the sign of the derivative f '(x) in each interval, as indicated in the figure. + + 0 /3 4/3 The sign scheme in the diagram indicates that we have relative minima at æ p 1 ç,- ö and æ 4 p 1 ç,- ö and a è 3 4ø è 3 4ø relative maximum at (, 0). We plot these points on a coordinate plane and place "cups" at the relative minima and a "cap" at the relative maximum. Finally, we sketch a smooth curve through these points, obtaining the graph shown in the figure. If the curve is defined over a particular interval, then one should also plot the endpoints. Find the points of maxima/minima of f(x) = x 3 1x and also draw the graph. f(x) = x 3 1x f(x) = 3(x 4) = 3(x ) (x + ) f(x) = 0 x = ± For tracing the graph let us find the maximum and minimum values of f(x). f( ) = 16, f() = 16. Find the values of a for which all roots of the equation 3x 4 + 4x 3 1x + a = 0 are real and distinct. Consider the function f(x) = 3x 4 + 4x 3 1x + a. Then f(x) = 1(x 3 + x x) = 1x (x 1) (x + ). From the sign scheme for f(x), we can see that the shape of the curve will be as shown below. Sketch the graph of f(x) = x 1/3 (x 4). f(x) = x 4/3 4x 1/3 4 1/3 4 /3 4 /3 f'(x) = x x x (x1) 3 3 3 We see that there is a relative minimum at (1, 3) and no extremum at (0, 0). We plot the critical points on a coordinate plane with a "cup" at (1, 3). Finally, we pass a smooth curve through these points, obtaining the graph shown in the figure. Note that at (0, 0) the curve has a vertical tangent. For four real and distinct roots, the two minima must lie below the x-axis and the maxima must lie above the x-axis. Thus, we have f( ) < 0 i.e. 48 3 48 + a < 0...(1) i.e. a < 3 and f(1) < 0 i.e. 3 + 4 1 + a < 0...() i.e. a < 5 and f(0) > 0 i.e. a > 0...(3)
MAXIMA AND MINIMA 7.19 Taking intersection of inequalities (1), () and (3) we have a (0, 5). Let f(x) = x 3 + 3(a 7) x + 3(a 9) x 1. If f(x) attains maxima at some positive value of x, then find the possible values of a. We have f(x) = x 3 + 3 (a 7)x + 3(a 9) x 1 and f(x) = 3x + 6(a 7) x + 3 (a 9) There are two critical points. Let, be roots of f(x) = 0 and let be the smaller root. Examining sign change of f(x). + + Also, we have f( ) = and f() = From the above facts, the graph of the curve y = f(x) can be drawn as shown below. Thus, if the maxima occurs at some positive value of x, then the minima must also occur at some positive value of x. This basically implies that both of roots f(x) = 0 must be positive and distinct, which it possible if discriminant D > 0 (a 7) > a 9 14 a + 58 > 0 a < 9/7...(1) and product of the roots > 0 1. Find the extrema of the function y = x x and draw its graph.. Find the points of local minima of the function, f(x) = 4x 3 x x, x [0, 3]. 3. Find the values of a so that f(x) = e x + (6 a)e x + 6ax has all its point(s) of extremum in (0, 1). 4. Determine the points of extrema of f(x) = x a 3 + (x b) 3 B A > 0 a 9 > 0 a < 3 or a > 3...() and sum of the roots > 0 a 7 < 0 a < 7...(3) Drawing the number line for inequalities (1), (), (3) and taking intersection, gives 9 a (, 3) 3, 7. For what values of a the point of local minima of f(x) = x 3 3ax + 3(a 1)x + 1 is less than 4 and point of local maxima is greater than. f(x) = 3(x ax + a 1) Clearly roots of the equation f(x) = 0 must be distinct and lie in the interval (, 4) D > 0 a R...(1) f( ) > a + 4a + 3 > 0 a < 3 or a > 1...() f(4) > 0 a 8a + 15 > 0 a > 5 or a < 3...(3) and < B A < 4 < a < 4 From (1), () and (3) 1 < a < 3. Alternative: f(x) = 3(x (a 1)) (x (a + 1)) Clearly < a + 1 < 4 and < a 1 < 4. 1 < a < 3. B 5. Prove that the function f(x) = 3x 4 4x 3 + 6x + ax + b has only one local minimum. 6. Find the least integral value of a for which the a 3 function f(x) = x + (a + )x + (a 1)x + 3 possess a negative point of local minimum. 7. For what real values of a and b are all the extreme values of the function, f(x) = a x 3 + ax x + b negative and the local maximum is at the point x 0 = 1?