Mathematic 108, Fall 2015: Solutions to assignment #7
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1 Mathematic 08, Fall 05: Solutions to assignment #7 Problem # Suppose f is a function with f continuous on the open interval I and so that f has a local maximum at both x = a and x = b for a, b I with a < b Explain why there must be value c (a, b) so f (c) 0 (Hint: If f (x) < 0 for all x (a, b), then any critical number of f in (a, b) is a local maximum) Solution # First, note that, since f has local maxima at x = a and x = b, it follows that there is at least one x in (a, b) where f(x) f(a) and at least one x in (a, b) where f(x) f(b) We know that, if we restrict f(x) to the interval [a, b], that there is a global minimum of f restricted to that region, and that, from before, there can t be an exclusive global minimum at an endpoint Therefore, the function must have a local minimum on the interior (ie, in the interval (a, b)) At this point, since f is twice differentiable, f (x) 0 (since if it was less than 0, the point would be a local maximum instead of a local minimum) Problem # Give an example of a continuous function with domain [, ] with a local maximum, but no local minimum Solution # In order for this to be true, since the function is continuous on a closed bounded set, the function must attain its absolute minimum on the endpoint One such function is f(x) = x (restricted to the appropriate domain), which has a local maximum at 0 Problem #3 Give an example of a function f with continuous second derivative for which f is zero at some point and whose graph does not have an inflection point Solution #3 x 4 is probably the simplest such function (note that in order for this to be true at a point x, f (x) must either equal 0 or not exist; otherwise, f (x) will be positive on one side of the zero and negative on the other side) Alternatively, f(x) = C, where C is constant, gives a trivial example Problem #4 Determine whether the following functions have an absolute maximum value and absolute minimum value on the given domain If it does determine the value a) f(x) = +e x on D = (, ) b) f(x) = tan (x) + +x on D = (, ) c) f(x) = x x + 3 on D = [0, ) Solution #4 a) We use the quotient and chain rules to evaluate f (x): f (x) = xe x ( + e x ) Since the denominator is always positive, as is e x, it holds that f (x) is positive when the numerator (and therefore when x) is positive, and negative when x is negative Therefore, there is a minimum at x = 0
2 b) By the chain and quotient rules again, we see that f (x) = + x x ( + x ) = x x + (x + ) = (x ) (x + ) This is equal to 0 when x = and positive otherwise Therefore, the only critical point, x =, is neither a local minimum nor a local maximum Since this is the only candidate, there are no global minima or maxima c) The derivative of this function is (x/ x + 3) This is always positive, since for x > 0, it follows that x = x < x + 3 (so x/ x + 3 is strictly less than ) Therefore, the function is always increasing, so the only candidate for a minimum or maximum is at the endpoint x = 0 Since the function increases to the right of 0, f(x) has a minimum at 0 Problem #5 Let f(x) = { x sin(/x) x 0 0 x = 0 and g(x) = sin(x) a) Use the it laws to show that f(x) g(x) = 0 (Hint: Consider f(x)/x g(x)/x ) b) Determine f (x) g (x) how do you reconcile this with a) and L Hospital s Rule Solution #5 a) As in previous homework assignments, we use the squeeze theorem and the inequality x x sin(/x) x to show that f(x)/x = 0 Additionally, sin(x)/x = is an established it Therefore, g(x) = f(x)/x g(x)/x = f(x) 0 b) f (x), by the product and chain rules, is equal to x sin(/x) cos(/x) whenever x 0 g (x) = cos(x) The numerator doesn t have a it as x goes to 0, so this it doesn t exist This implies that f (x) g (x) isn t a continuous function (formally, this breaks the requirement that f (x) g (x) exists) Problem #6 Use L Hospital s Rule to evaluate the following its arcsin(x) a) x ( ) b) + x arctan(x) c) + ( + sin(x)) /x Solution #6 a) This is a 0/0 type it, so we can use L Hospital s Rule directly By the chain rule, the derivative of the numerator is and the derivative of the denominator is This quotient, evaluated at x = 0 is equal to (x)
3 b) This isn t a 0/0 it as written, so we need to use the equivalent expression + arctan(x) x x arctan(x) This is 0 0, so we can use L Hospital s Rule If f(x) = arctan(x) x and g(x) = x arctan(x), then f (x) = +x and g (x) = arctan(x) + x x + Both of these are 0 at 0, so we can apply L Hospital s Rule again f (x) = x (x +) and g (x) = +x + +x x f (+x ) (0) g (0) = 0, so this it is 0 c) This is a type it Therefore, we can take the natural log and evaluate the it of that, then convert it back to our original it by exponentiating We know ln(( + sin(x)) /x ) = ln(+sin(x)) x This is a proper 0/0 it, so we can now apply L Hospital s rule Given f(x) = ln( + sin(x)) and g(x) = x, it follows from the chain rule that f (x) = cos(x) +sin(x) and g (x) = Evaluating this directly, we see that at 0 the quotient of these two (and therefore the log of the original it) is equal to Therefore, the value of the original it is e Problem #7 Suppose f is differentiable, f(3) = and f (3) = Evaluate f(3+x) f(3 4x) x Solution #7 This is a 0/0 it, so we can use L Hospital s rule By the chain rule: f(3 + x) f(3 4x) = f (3 + 0) + 4f (3 + 0) x = 0 Problem #8 Suppose g is differentiable, g() = 0 and g () = 3 Evaluate x g(+x)+g(4 x) x Solution #8 For this problem, we can evaluate directly (because all functions involved are continuous and the denominator is nonzero) Plugging in x =, this expression evaluates to 0 Problem #9 Use the methods of Section 45 to sketch the following curves a) y = (x ) x + b) y = x sin(x) c) y = + x x Solution #9 Sketches are left to the reader s imagination, or alternatively, graphing software a) Domain: Since x + is positive, the domain is all real numbers Intercepts: y(0) = 4, y = 0 at x = (since the numerator is 0 then) Symmetry: This function has none Asymptotes: Since the function is continuous, there are no vertical asymptotes Since y = x 4x+4 x +, and the it of this as x goes to ± is either way, there is a horizontal asymptote at y = Intervals of Increase or Decrease: y = (x +)(x 4) (x 4x+4)(x) (x +), which can be simplified to (x 3 4x +x 4) (x 3 8x +8x) (x +), and subsequently to 4x 6x 4 (x +) The numerator factors into (x )(4x+) This is positive for x < and x >, and negative between the two roots
4 Local Maximum and Minimum Values: From the previous calculations, there is a local max at and a min at Concavity and Points of Inflection: The second derivative of this doesn t have roots that are easy to deal with, so looking at convexity and concavity wouldn t be particularly enlightening beyond the information we can derive from the increase/decrease behavior and asymptotes b) Domain: The domain of this function is all real numbers Intercepts: y(0) = 0 Other than that, other intercepts are too difficult to find Symmetry: This is an odd function Asymptotes: This function has no asymptotes, due to the oscillatory nature of sine Intervals of Increase or Decrease: y = cos(x) This is positive whenever cos(x) < and negative when cos(x) > Therefore, x must be in the interval (( ( ) ) k 6) π, k + 6 π in order for y to be negative, where k is any integer, and in (( ( ) k + 6) π, k + 5 6) π in order for y to be positive Local Maximum and Minimum Values: As before, the derivative is periodic, so there are infinitely many points where y = 0 By the previous part, ( k + 6) π is a local min, where k ranges over the integers, and ( k 6) π is a local max Concavity and Points of Inflection: y = 4 sin(x) This is positive (and therefore y is concave up) when k is between ( kπ, ( ) (( ) k + ) π and negative (so y is concave down) between k + ) π, (k + )π Therefore, inflection points are at kπ and ( k + ) π c) Domain: The domain of this function is all real numbers Intercepts: As + x > x, y is always going to be positive Symmetry: This has no symmetry Asymptotes: This has a horizontal asymptote of x = 0 as x goes to (to see this, use rational conjugation to rewrite y as x ++x ) Since we have that x x + x = 0, it follows that x x + x = 0, since this function is even and behaves like our original function for x > 0, and therefore x x + + x = 0 (ie, this shows us that x + has an oblique, or slant, asymptote of y = x as x goes to This tells us that x ( x + x) ( x) = 0, so y has a slant asymptote of x Intervals of Increase or Decrease: y = function is always decreasing x +x, which is necessarily less than 0 Therefore, the Local Maximum and Minimum Values: There are no local minima or maxima Concavity and Points of Inflection: y = +x, or equivalently, The latter is (+x ) 3/ (+x ) 3/ always positive, so the function is always concave up x(x) Problem #0 Consider the family of polynomials P c (x) = x 3 + 3cx + 3x a) Determine the values of c so that P c has both a local maximum and a local minimum b) Sketch the graph of y = P c (x) for a value c for which c has both a local maximum and minimum and sketch the graph for a value c for which it does not Solution #0 a) P c(x) = 3x + 6cx + 3 = 3(x + cx + ) In order to find the critical points, we apply the quadratic formula to x + cx + (or the original 3x + 6x + 3) in order to find roots at c± 4c 4 The number of roots therefore depends on the behavior of the discriminant 4c 4: if it s positive, there are two distinct roots, if it s 0, there s only one root, and if it s negative, there are no real roots Therefore,
5 in order to have a local minimum and a local maximum, there must be two distinct critical points, so c > or c < When this happens, by the properties of quadratic equations, P c(x) is positive to the left of the first (leftmost) root, negative between the two roots, and positive to the right of the second root Therefore, there is a local maximum at c c and a local minimum at c + c b) You can graph these yourselves, A case for which a local maximum and minimum exist is c = and a case for which they don t is c = 0 It might also be interesting to graph the border cases c = ±, in which f has a critical point which is neither a minimum nor maximum (in fact, each of these cases are translations of the graph of y = x 3, which has the same properties) Book Problems a) Section 44: #4, #3, #44, #76, #88 b) Section 45: #, #, #50, #7 c) Section 46: #8 Solution #444 Only (b) and (e) fail to be indeterminate forms, with its 0 and respectively Solution #443 This is / The derivative of the numerator is ln(x)/x and the derivative of the denominator is, so the quotient, ln(x)/x, once again requires L Hospital s Rule The derivative of the numerator is /x and the derivative of the denominator is The quotient here goes to 0 as x goes to infinity Solution #4444 This is a 0 case, so we can rewrite it as x Applying L Hospital s Rule gives e x/ us xe x/, which goes to 0 sec x tan x Solution #4476 The first application of L Hospital s rule gives us sec (x), or tan x sec x The second application brings us back to where we started, and further applications keep alternating between the two If we write the quotient as the equivalent expression sin x, the it is easily shown to be Solution #4488 We start off by writing this as ( sin(x) + ax 3 ) + bx x 3 This is 0/0, so we use L Hospital s Rule to get ( cos(x) + 3ax ) + b 3x In order for this it to have a chance of existing, since the denominator is 0, the numerator must be 0 as well Therefore, b = Talking L Hospital s Rule, we next get ( ) 4 sin(x) + 6ax 6x and, once more,
6 ( ) 8 cos(x) + 6a 6 Therefore, in order for the it to be 0, a = 4/3 Solution #45, 45, 4550 These use techniques similar to 9, so for the sake of getting the solutions finished before the midterm, I won t go through all of the details, but notable features include a vertical asymptote at x = 0 for problem, and a restricted domain of (, ) (and a corresponding vertical asymptote at the left endpoint) for problem 50 Solution #457 The y = x+ slant asymptote is equivalent to saying that x x + 4x (x+) = 0 Note that we chose positive infinity here from cursory examination of the behavior of x + and x + 4x, which both go to infinity as x goes to infinity; if x goes to, the two functions blow up in different directions We can multiply by the conjugate x +4x+(x+) x +4x+(x+) to get 4 x, which goes to 0, so x + +4x+(x+) is a slant asymptote The proof of the other slant asymptote proceeds similarly, except to The graph can be drawn from this behavior, as well as first and second derivative behavior Solution #468 This is similar to Problem 0 from the worksheet, where there are maxima and minima when c < 0, and none otherwise (but a critical point when c = 0)
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