MEAN VALUE THEROEM

Let f be a function that satisfies the following hypotheses:

1)  f is continuous on the closed interval [a, b]

2)  f is differentiable on the open interval (a, b)

Then there is a number c in (a, b) such that

f'(c)  =  [f(b) - f(a)] / (b-a)

f'(c)(b-a)  =  [f(b) - f(a)]

the Mean Value Theorem, in the form given by 

f'(c)  =  [f(b) - f(a)] / (b-a)

that there is at least one point P(c, f (c)) on the graph where the slope of the tangent line is the same as the slope of the secant line AB

Use of Mean value Theorem :

There is a point P where the tangent line is parallel to the secant line AB. 

Verify that the function satisfies the hypotheses of the Mean Value Theorem on the given interval. Then find all numbers c that satisfy the conclusion of the Mean Value Theorem.

Example 1 :

f(x)  =  2x2-3x+1, [0, 2]

Solution :

(i)  f(x) is defined and continuous on [0, 2]

(ii)  f(x) is differentiable on the interval (0, 2).

(iii)  f(x)  =  2x2-3x+1

f'(x)  =  4x-3

f'(c)  =  4c-3

f(a)  =  f(0)  ==>  2(0)2-3(0) + 1  ==>  1

f(b)  =  f(2)  ==>  2(2)2-3(2) + 1  ==>  3

f'(c)  =  [f(b) - f(a)] / (b-a)

4c-3  =  (3-1)/(2-0)

4c-3  =  1

c  =  1 ∈ [0, 2]

Example 2 :

f(x)  =  x3-3x+2, [-2, 2]

Solution :

(i)  f(x) is defined and continuous on [-2, 2]

(ii)  f(x) is differentiable on the interval (-2, 2).

(iii)  f(x)  =  x3-3x+2

f'(x)  =  3x2-3

f'(c)  =  3c2-3

f(a)  =  f(-2)  ==>  (-2)3-3(-2)+2 

f(-2)  =  -8+6+2

f(-2)  =  0

f(b)  =  f(2)  ==>  23-3(2)+2 

f(2)  =  8-6+2

f(2)  =  4

f'(c)  =  [f(b) - f(a)] / (b-a)

3c2-3  =  (4-0)/(2+2)

3c2-3  =  1

3c2  =  4

c2  =  4/3

c  =  √(4/3)

c  = (±2/√(3)

Example 3 :

Find the number c that satisfies the conclusion of the Mean Value Theorem on the given interval. Graph the function, the secant line through the endpoints, and the tangent line at (c, f(c)). Are the secant line and the tangent line parallel?

(i)  f(x)  =  √x,  [0, 4]

(ii)  f(x)  =  e-x, [0, 2]

Solution :

(i)  f(x)  =  √x,  [0, 4]

f'(x)  =  1/2√x

f'(c)  =  1/2√c  ---(1)

f'(c)  =  [f(b) - f(a)]/(b-a)

f(b)  =  f(4)  =   √4  =  2

f(a)  =  f(0)  =   √0  =  0

[f(b) - f(a)]/(b-a)  ==>  (2-0)/(4-0)

=  2/4

=  1/2  ---(2)

(1)  =  (2)

1/2√c  =  1/2

√c  =  1

c  =  1

(ii)  f(x)  =  e-x, [0, 2]

f'(x)  =  -e-x

f'(c)  =  -e-c  ---(1)

f'(c)  =  [f(b) - f(a)]/(b-a)

f(b)  =  f(2)  =   e-2

f(a)  =  f(0)  =    -e-0  =  1

[f(b) - f(a)]/(b-a)  ==>  (e-2-1)/(2-0)

=  (e-2-1)/2  ---(2)

(1)  =  (2)

-e-c  =  (e-2-1)/2

-2e-c  =  e-2-1

e-c  =  (1-e-2)/2

-c  =  ln[(1-e-2)/2]

c  =  0.8386

Example 4 :

The graph of a function f is shown. Verify that f satisfies the hypotheses of Rolle’s Theorem on the interval [0, 8]. Then estimate the value(s) of c that satisfy the  conclusion of Rolle’s Theorem on that interval.

Solution :

(1) f is continuous on the closed interval [0, 8] .

(2) f is differentiable on the open interval (0, 8) .

(3) f (0) = 3 and f(8) = 3

Thus, f stratifies the hypotheses of Rolle’s Theorem. The numbers c = 1 and c = 5 satisfy the conclusion of Rolle’s Theorem

since f'(1)  =  f'(5)  =  0.

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