Test 3 Review. y f(a) = f (a)(x a) y = f (a)(x a) + f(a) L(x) = f (a)(x a) + f(a)

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1 MATH 2250 Calculus I Eric Perkerson Test 3 Review Sections Covered: 3.11, Topics Covered: Linearization, Extreme Values, The Mean Value Theorem, Consequences of the Mean Value Theorem, Concavity and Curve Sketching, Optimization, L Hôpital s Rule Linearization Given a function f(x) and a point x = a, linearization is a technique that uses the tangent line to the graph of f(x) at the point a as a linear approximation to the function. To find the linearization, L(x), we start with the tangent line at a and rewrite it as a function: y f(a) = f (a)(x a) y = f (a)(x a) + f(a) L(x) = f (a)(x a) + f(a) This is called linearization because the result L(x) is a line. It is important because L(x) is in fact the best linear approximation to the function f(x). Example: Find the linearization of f(x) = x at the point a = 9. Use this linearization to estimate the value of 8. The linearization is given by L(x) = d dx ( x) x=9 (x 9) + 9 = (x 9) + 9 = 1 (x 9) we can now use L(8) = 1 6 (8 9) + 3 = = 17 6 = as an approximation for finding 8 = Notice how good of an approximation this is: the error is only = x L(x) = (1/6)(x - 9)

2 4.1 Extreme Values A function f(x) has an absolute maximum at a point c if the value of f(c) is greater (or equal to) than all other values of f(x). To be precise, we say: Definition: A function f(x) with domain D has an absolute (or global) maximum at a point c if f(c) f(x) for all x in D. Similarly, a function f(x) with domain D has an absolute (or global) minimum at a point c if f(c) f(x) for all x in D. Not every function will have an absolute maximum or an absolute minimum. A good example is f(x) = x 3. However, the Extreme Value Theorem lets us know that certain functions will always have both an absolute maximum and an absolute minimum. Theorem: (The Extreme Value Theorem) If f(x) is a continuous function on a closed interval [a, b], then f(x) has both an absolute maximum and an absolute minimum. Note that the Extreme Value Theorem doesn t say that continuous function on non-closed intervals or that discontinuous functions don t have absolute extrema (maximums or minimums), just that we don t know on this basis alone. Global extrema (maximums and minimums) can sometimes be hard to find. However, it turns out that for differentiable functions it is relatively easy to find points where the function has a local extrema. Because any global extrema must also be a local extrema, the derivative is a valuable tool in telling us where to look for absolute extrema. We define local extrema as follows: Definition: A function f(x) has a relative (or local) maximum at a point c if f(c) f(x) for all x near the point c, i.e. in a small open interval around c. Similarly, a function f(x) has a relative (or local) minimum at a point c if f(c) f(x) for all x near the point c. Using the derivative gives us the following theorem, one of the most important in calculus: Theorem: (The First Derivative Theorem for Local Extrema) If f(x) has a local extremum at an interior point (i.e. not an end point) of its domain, and if f(x) is differentiable at c (i.e. f (c) exists), then f (c) = 0. This tells us that if we re looking for local extrema, then we only have to look at where either: f (x) = 0 at an interior point, or where f (x) is undefined at an interior point, or where x is an end point of the domain. The following definition is helpful in this regard: Definition: A point x where f (x) is undefined or where f (x) = 0 is called a critical point. 2

3 Notice in the above picture that all of the local maximums and local minimums (including the global ones, of course, because any global maximum is also a local maximum, and similarly for minima) occur at either end points, or critical points. This is so important it needs to be said again: Theorem: Any local minima or maxima of a continuous function f(x) must occur at either critical points or end points. However, we still aren t guaranteed that a global maximum or minimum occurs, unless our function is continuous and defined on a closed interval, in which case the Extreme Value Theorem tells us so. This breaks our approach into two pieces: Technique to Find Global Extrema: If f(x) is continuous on a closed interval: Compute f(x) at all critical points and both end points. The global maximum is the largest of these values and the global minimum is the smallest of these values. If f(x) is continuous but not defined on a closed interval: Compute f(x) at all critical points, and if there is an endpoint, compute f(x) there as well. The largest of these values is the only possible candidate for the global maximum, but it is not necessarily the global maximum (similarly for minima). If it is not the global maximum (minimum), you need prove it by finding a point larger (smaller) than that point, or by showing that at some point a, maybe infinity, that lim x a f(x) = (or ). It it is the global maximum (minimum), you need to prove that the function never get larger (smaller) anywhere on the domain. This is done by looking at where the function is increasing and where the function is decreasing Example: Find the global maximum and minimum of the function f(x) = x 2/3 on the interval [ 2, 3]. Because this function is continuous on a closed interval, we compute f(x) at all critical points and both end points. To find the critical points, we need to know where f (x) either does not exist or where f (x) = 0. So we compute f (x) = d dx x( 2/3) = 2 3 x 1/3 = 2 3x 1/3 notice that this derivative does not exist for x = 0, but otherwise exists and is not zero. So x = 0 is the only critical point. So we have to compute f(x) at the two endpoints and at the critical point x = 0. x f(x) Comparing the values shows that the global max is 3 9 and occurs at x = 3, and that the global min is 0 and occurs at x = x 2/

4 Example: Find the global maximum and minimum, if they exist, of the function g(x) = e x2. The function g(x) is continuous, but the domain of the function is the set of all real numbers, (, ), which is not a closed interval. In this case we need to compute the function at all critical points (in this case we don t have any end points). So we find g (x) = d 2 dx e x = 2xe x2 Because e y is always positive, e x2 is always positive. Therefore, we have that, if x < 0, then g (x) > 0 and hence g(x) is increasing; and if x > 0, then g (x) < 0 and hence g(x) is decreasing. Because g(x) is always increasing to the left of x = 0, and always decreasing afterwards, the function does have an absolute max at x = 0, and that absolute max is g(0) = 1. Because g(x) is always increasing to the left of x = 0, and always decreasing afterwards, we also know that g(x) cannot have an absolute minimum. e -x Mean Value Theorem Theorem: (Rolle s Theorem) If f(x) is continuous on [a, b] and differentiable on (a, b), and if f(a) = f(b), then there exists at least one point c in the interval (a, b) at which f (c) = 0. Theorem: (Mean Value Theorem) If f(x) is continuous on [a, b] and differentiable on (a, b), then there exists at least one point c in the interval (a, b) at which f (c) = f(b) f(a) b a (Note that f(b) f(a) b a is the slope of the line connecting the points (a, f(a)) and (b, f(b))). Corollary: If f (x) = 0 for all x in the interval (a, b), then f(x) = C for all x in (a, b), where C is a constant. Corollary: If f (x) = g (x) for all x in the interval (a, b), then f(x) = g(x) + C for all x in (a, b), where C is a constant. Note that the second corollary follows from the first when we write the function h(x) = f(x) g(x). Then if f (x) = g (x), we know that h (x) = f (x) g (x) = 0 and from the first corollary, we have that h(x) = C, and therefore f(x) g(x) = C, and hence f(x) = g(x) + C 4

5 Example: Suppose that f (x) = 3x and that f(0) = 4. Find f(x). We want to think of another function g(x) that has g (x) = f (x) = 3x. You can check that g(x) = 3 2 x2 is such a function. Then the second corollary to the mean value theorem tells us that f(x) = g(x)+c = 3 2 x2 +C for some unknown constant C. Using our additional piece of information, i.e. that f(0) = 4, we have that 4 = 3 2 (0)2 + C. Therefore, C = 4, and so f(x) = 3 2 x First Derivative Test A third corollary to the Mean Value Theorem is the following: Corollary: If f (x) > 0 for all x in the interval (a, b), then f(x) is increasing on [a, b]. Theorem: f(x) changes from increasing to decreasing at a point, i.e. if f (x) changes from positive to negative at a a point, then f(x) has a local maximum at that point. If f(x) changes from decreasing to increasing at a point, i.e. if f (x) changes from negative to positive at a a point, then f(x) has a local minimum at that point. If f (x) does not change sign at a point, then that point is not a local extremum. 4.4 Concavity and Curve Sketching Definition: The graph of a differentiable function f(x) is concave up on an open interval if f (x) is increasing on that interval. The graph of a differentiable function f(x) is concave down on an open interval if f (x) is decreasing on that interval. It s important to note that concavity of a function f(x) is independent of whether or not f(x) is increasing or decreasing. The four different combinations of increasing/decreasing and concave up/concave down are shown below. From what we know about increasing and decreasing functions from the last section, we know that f(x) is increasing (decreasing) if f (x) > 0 (< 0). Because the second derivative f (x) is just the derivative of the first derivative f (x), we have the following: 5

6 Theorem: (The Second Derivative Test for Concavity) Let f(x) be a twice differentiable on an interval I. Then If f (x) > 0 on I, the graph of f(x) is concave up. If f (x) < 0 on I, the graph of f(x) is concave down. Definition: A point where the graph of f(x) has a tangent line and where the concavity changes is called a point of inflection. Example: The function f(x) = x 3 has a point of inflection at x = 0, as the second derivative f (x) = 6x changes from negative to positive at x = 0, and so by the second derivative test for concavity, the concavity of the graph of f(x) changes at x = 0. And because f(x) does have a tangent line at x = 0, x = 0 is a point of inflection x 3-5 Note that at a point of inflection, either f (x) does not exist or f (x) = 0. However, this is not sufficient for a point to be a point of inflection. Example: The function f(x) = x 4 does not have a point of inflection at x = 0, as the second derivative f (x) = 12x 2 is positive to the left of x = 0 and is still positive to the right of x = x

7 Technique for Sketching Graphs: Identify the domain of f(x) and any symmetries it might have. Find f (x) and f (x). Find the critical points of f(x), and identify the function s behavior at each one (i.e, is the point a local maximum, minimum, or neither). Find where f(x) is increasing and decreasing. Find points of inflection and determine the concavity of f(x). Identify any asymptotes. Plot key points, such as the intercepts and any points found in steps 3 5, and sketch the curve. 4.5 L Hôpital s Rule Indeterminate Forms: Expressions of the form 0 0,, 0,, 1, 0 0, 0. These expressions are not numbers, and hence can t be equal to a number. Theorem: (L Hôpital s Rule) If either or then so long as all of these expressions exist. lim f(x) = lim g(x) = 0 x c x c lim f(x) = lim g(x) = x c x c f(x) lim x c g(x) = lim f (x) x c g (x) Sometimes when using L Hôpital s Rule, you will have to use the rule more than once, and sometimes you have to rewrite the expression you are working with in order to get an expression where you can use L Hôpital s Rule. Example: lim x x sin( 1 x ) First we need to rewrite this expression to use L Hôpital s Rule. Now notice that lim x sin( 1 x x ) = lim sin( 1 x ) x 1 x lim sin( 1 x x ) = sin(0) = 0 and that 1 lim x x = 0 so in fact we can use L Hôpital s Rule. So we have that sin( 1 x lim ) x 1 = lim x x d dx sin( 1 x ) d 1 dx x cos( 1 x = lim )( x 2 ) x x 2 = lim cos( 1 x x ) = cos(0) = 1 7

8 4.6 Optimization Don t forget when doing optimization problems that you have do Demonstrate that a point is a local extremum. Demonstrate that a point is a global extremum (This always requires you to say something about the domain of your function, and not just the domain of the function you have, but the effective domain, i.e. the domain that makes sense given the model you are working with). Answer in a sentence with units. 8