1. Mean Value Theorem
for
Derivatives

2. Mean Value Theorem for Derivatives
If f (x) is continuous over [a,b] and differentiable over (a,b), then at some point c between a and b:
Mean Value Theorem for Derivatives
The Mean Value Theorem says that at some point in the closed interval, the actual slope equals the average slope.
The Mean Value Theorem only applies over a closed interval.

3. Tangent parallel to chord.
Slope of tangent:
Slope of chord:

4. A couple of somewhat obvious definitions:
A function is increasing over an interval if the derivative is always positive.
A function is decreasing over an interval if the derivative is always negative.

5. These two functions have the same slope at any value of x.
Functions with the same derivative differ by a constant.

6. Find the function whose derivative is and whose graph passes through .
Example 6:
Find the function whose derivative is and whose graph passes through
so:
could be
or could vary by some constant .

7. Find the function whose derivative is and whose graph passes through .
Example 6:
Find the function whose derivative is and whose graph passes through
so:
Notice that we had to have initial values to determine the value of C.

8. This section is just an introduction.
The process of finding the original function from the derivative is so important that it has a name:
Antiderivative
A function is an anti-derivative of a function
if for all x in the domain of f. The process of finding an anti-derivative is anti-differentiation.
You will hear much more about antiderivatives in the future.
This section is just an introduction.

9. Example 7b: Find the velocity and position equations for a downward acceleration of 9.8 m/sec2 and an initial velocity of 1 m/sec downward.
(We let down be positive.)

10. Example 7b: Find the velocity and position equations for a downward acceleration of 9.8 m/sec2 and an initial velocity of 1 m/sec downward.
The power rule in reverse: Increase the exponent by one and multiply by the reciprocal of the new exponent.
Since velocity is the derivative of position, position must be the anti-derivative of velocity.

11. Example 7b: Find the velocity and position equations for a downward acceleration of 9.8 m/sec2 and an initial velocity of 1 m/sec downward.
The initial position is zero at time zero. p

12. ASSESSMENT
Mean Value Theorem

13. Que 1:
State and proov Mean Value Theorem.
Que 2:
Verify Mean Value Theorem for the function f(x) = -2x 3 + 6x – 2 in the interval [– 2 , 2] and find the  value of c .
Find the value of c that satisfies the conclusion of the mean value theorem for f (x) = x3 - 2x2 - x + 3 on[0, 1].
Que 3:
Find the value of c that satisfies the conclusion of the mean value theorem for the function f (x) = ln x in[1, e].
Que 4:

14. SOLUTIONS
Solution 2:
Solution 3:
ln (e - 1)
Solution 4:

15. Multiple Choice Question

16. Que 1: We verify the "Mean Value theorem" for a function f (x):
Que 2: State whether the function f (x) = 3x2 - 2 on [2, 3]satisfies the mean value theorem.
Que 3: Which of the following functions satisfies the mean value theorem?

17. Que 4. Determine the point on the parabola f (x) = (x - 2)2, at which the tangent is parallel to the chord joining the points (2, 0) & (3, 1). 
 Que 5. Find the point on the parabola y = (x + 3)2, at which the tangent is parallel to the chord of the parabola joining the points (- 3, 0) & (- 4, 1).

18. Ques 7. State whether the function f (x) = sin x - sin 2x, x ∈ [0, π] satisfies the mean value theorem.
Que 6. State whether the function f(x) = ln x on[1, 2] satisfies the mean value theorem.

19. Solutions
Ans 1.
d. If f(x) is continuous on [a, b] and is differentiable on (a, b), then only it follows the hypotheses of mean value theorem for which the mean value theorem can be verified
Ans 2.
b. f (x) = 3x2 - 2, x ∈ [2, 3] [Given function.] The given function is continuous on [2, 3] and differentiable in (2, 3). So, there exists c ∈ (2, 3) such that f ′ (c) =f(3) - f(2)3 - 2. [By mean value Theorem.] f ′(c) = 15. [f ′(c) = 6c. ] 6c = 15 ⇒ c = 5 / 2 ∈ (2, 3). Hence, the given function satisfies the mean value theorem.

20. c. f (x) = x2 - 6, the function f (x) is continuous on [7, 8] and differentiable on (7, 8) 
[Consider the choice A.] f ′ (c ) = f (8) - f (7)8 - 7 = 15 [Try to get c ∈ (7, 8) such that f ′ (c ) =f (b) - f (a)b - a] 2c = 15 [f ′ (x) = 2x]  c = 15 / 2 ∈ [7, 8] [There exists c ∈ (7, 8) such that f ′ (c ) =f (b) - f (a)b - a] So, the function in the choices A satisfies the mean value theorem f (x) = log x, x ∈ [- 7, 8] is not continuous in ∈ [- 7, 8] and hence does not satisfy the mean value theorem.
[Consider the choice B.]
Ans 3.

21. f (x) = [x], x ∈ [- 7, 7] is not continuous in [- 7, 7] and hence does not satisfy the mean value thoerem
[Consider the choice C.] f (x) = |x|, x ∈ [- 8, 8] is continuous in [- 8, 8] but not differentiable in (- 8, 8) and hence does not satisfy the mean value theorem . [Consider the choice D.]
a. Slope of the chord joining (2, 0) & (3, 1) = 1-0 / 3-2 = 1 Slope of the tangent to the curve at any point (x, f(x)) is f ′ (x) = 2(x - 2) f ′ (c) = 1 [By Mean Value Theorem.]  2(c - 2) = 1 c = 5 / 2 ∈ (2, 3) f (c) = (c - 2)2 = (5 / 2 - 2)2 = 1 / 4 Hence, the point where the tangent to the parabola is parallel to the given chord is (c, f (c)) = (5 / 2, 1 /4)
Ans 4.

22. Ans 5. d.
Slope of the chord joining the points (- 3, 0) & (- 4, 1) = (1 -  0)( - 4 + 3) = - 1  Slope of the tangent to the curve at any point (x, f(x)) is f ′ (x ) = 2(x + 3) f ′(c) = -1 [By mean value theorem.] 2(c + 3) = -1 c = - 7 / 2 ∈ (-4, - 3) f (c) = (c + 3)2 = (- 72 + 3)2 = 14. Hence, the point where the tangent to the parabola is parallel to the given chord is (c, f(c)) = (- 7 / 2, 1 /4). 
a. f(x) = ln x, x ∈ [1, 2] [Given function.] f(x) is continuous on [1, 2] and differentiable on (1, 2). So, there exists c ∈ (1, 2) such that f ′ (c) =f(2) - f(1)2 - 1. [By mean value Theorem.] 1c = ln 2 ⇒ c = 1ln 2 ∈ (1, 2). [f ′(x) = 1x] Hence, the given function satisfies the mean value theorem.
Ans 6.

23. a. f (x) = sin x - sin 2x, x∈ [0, π] [Given function
a. f (x) = sin x - sin 2x, x∈ [0, π] [Given function.] The given function is continuous on [0, π] and differentiable in (0, π). So, there exists c ∈ (0, π) such that f ′ (c) =f(π) - f(0)π  - 0. [By mean value theorem.] cos c - 2 cos 2 c = 0 [f ′ (x) = cos x - 2 cos 2x.] cos c = 4 cos2c - 2 [cos 2c = 2 cos2c - 1.]  4 cos2 c - cos c - 2 = 0 c = 32o.53′ , 126o.37′ ∈ (0, π) ) [Solve use the calculator.] Hence, the function f (x) satisfies the mean value theorem.
Ans 7.