Applications of Derivatives AP/Honors Calculus Chapter 4 Applications of Derivatives

Topics 4.1 Extreme Values 4.2 Mean Value Theorem 4.3 Connecting the Graph of f to f’ and f” 4.4 Modeling and Optimization 4.5 Linearization (and Newton’s Method) 4.6 Related Rates

The Mean Value Theorem What does this mean? If y = f(x) is continuous at every point on [a,b] and differentiable on every point on (a,b), then there is at least one point c on (a,b) such that What does this mean? It means that for a function that meets the given conditions: 1) there will be at least one point on [a,b] whose slope of its tangent is equal to the slope of the secant from a to b. 2) It means that there must be at least one point on [a,b] where the instantaneous rate of change is equal to the average rate of change on [a,b].

Some Definitions and Theorems Let f be a function defined on some interval R and let m and n be any two points on R, then 1) f increases on R if m < n ➾ f(m) < f(n) 2) f decreases on R if m < n ➾ f(m) > f(n) Let f be continuous on [a,b] and differentiable on (a,b). 1) If f’ > 0 for every point on (a,b), then f increases on [a,b] 2) If f’ < 0 for every point on (a,b), then f increases on [a,b] A function that either strictly increases or strictly decreases on an interval is said to be monotonic on that interval.

Some Definitions and Theorems 2 If f’(x) = 0 for each point of some interval R, then there is a constant C for which f(x) = C for all x in R. if f’(x) = g’(x) for each point of an interval R, then there is a constant C such that f(x) = g(x) + C for all x in R. A function F(x) is an antiderivative of f(x) if F’(x) = f(x) for all x in the domain of f. The process of finding the antiderivative is antidifferentiation.

Example 1 f(x) = 2x3 – x2 + 5 1) Find where f has local extrema. 2) Find the intervals where f is increasing or decreasing. (The intervals of monotonicity.)

Example 2 Show that the Mean Value Theorem is satisfied by f(x) = 2x2 – x – 3 on [–2, 4] Find the average rate of change on the interval Set the first derivative equal to the average rate of change on the interval f ’(x) = 4x – 1 Set 4x – 1 = 3 4x = 4 x = 1 Since 1 is on the interval [–2, 4], we have shown at least one point on the interval whose slope of its tangent is equal to the slope of the secant.

Example 3 Find all possible functions f with the given derivative f ‘(x) = 3x2 – 2x + 1 f(x) = x3 – x2 + x + C

Example 4 Sketch a possible graph of a differentiable function y = f(x) that has the following properties. f(3) = 4 f ‘(3) = 0 f ‘(x) > 0 for 0 < x < 3 f ‘(x) < 0 for x > 3 and for x < 0 f(0) = –3 f ’(0) = 0 f(–4) = –6 f ’(–4) = 0

Assignment Begin 4.2