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

2. 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

3. 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].

4. 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.

5. 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.

6. 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.)

7. 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.

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

9. 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

10. Assignment
Begin 4.2