1. 5-2 mean value theorem

2. *The Mean Value Theorem says that somewhere between A and B on a differentiable curve there is at least one tangent line parallel to the chord AB* B A a c b
Thm: Mean Value Theorem for Derivatives
If y = f (x) is continuous at every point of closed interval
[a, b] and differentiable at every point of its interior (a, b),
then there is at least one point c in (a, b) at which
Note: MVT tells us that c exists, but not how to find it

3. Ex 1) Show that the function f (x) = x2 satisfies the hypothesis of the Mean Value Theorem on the interval [0, 2].
Then find a solution c to the equation
on this interval.
Cont?
Diff ?
At x = 1, the slope is 2
want

4. Ex 2) Explain why each of the following functions fails to satisfy the conditions of the Mean Value Theorem on the interval [–1, 1]. a) b)
Cont?
Diff ?
x = 0, corner
Cont?
No… jumps at x = 1

5. Ex 3) Let
Find a tangent to f in the interval (–1, 1) that is parallel to the secant AB.
equation
Cont?
Diff ?
Pts:
A = (–1, 0)
B = (1, 0)
Parallel  same slope
–x = 0
x = 0
Point (0, 1) m = 0  y = 1

6. Def: Increasing and Decreasing Functions
If x1 < x2, then
f increases if f (x1) < f (x2)
f decreases if f (x1) > f (x2)
Cor: If f (x) is continuous on [a, b] and differentiable on
(a, b), then
If f ' > each point of (a, b), then f increases on [a, b]
(2) If f ' < each point of (a, b), then f decreases on [a, b]

7. Ex 5) Where does the graph of y = x2 rise and fall?
Look at deriv  y' = 2x
Dec:
2x < 0
x < 0
(–, 0)
Inc:
2x > 0
x > 0
(0, )

8. Ex 6) Where is the function f (x) = x3 – 4x increasing and where is it decreasing?
Look at deriv  f '(x) = 3x2 – 4
Dec:
3x2 – 4 < 0
3x2 < 4
Inc:
3x2 – 4 > 0
3x2 > 4
Dec:
Inc:

9. Cor: Functions with f '(x) = 0 are constant
Cor: Functions with the same derivative differ by a constant
If f '(x) = g'(x) at each point on interval I, then there is a constant C such that f (x) = g (x) + C for all x in I.
Ex 7) Find the function f (x) whose derivative is sin x and whose graph passes through the point (0, 2).
f '(x) = sin x  f (x) = – cos x + C
Know (0, 2)  f (0) = 2  2 = – cos (0) + C
2 = –1 + C
3 = C
f (x) = – cos x + 3

10. Def: Antiderivative
A function F (x) is an antiderivative of function f (x) if
F '(x) = f (x) for all x in the domain.
Note: The process to find F (x) is called antidifferentiation.

11. homework
Pg # 2, 10, 12, 18, 22, 28
Pg # 6-48 (mult of 6)