Solve the following Mean Value Problems

Question

1. Given f(x) = 3x^2 + 1, find c such that f(c) equals the average of f(x) =

3x^2 + 1 over (0,2).

2. Given f(x) = x^3 - 1, find c such that f(c) equals the average of f(x) = x^3 -

1 over (-2,4).

3. Given f(x) = sqrt(x-1) + 2, find c such that f(c) equals the average of f(x)

= sqrt(x-1) + 2 over (2,10)

.

4. Given f(x) = sqrt(x) - 2, find c such that f(c) equals the average of f(x)

= sqrt(x) - 2 over (0,1/2).

5. Given f(x) = 2x/3, find c such that f(c) equals the average of f(x) = 2x/3

over (-5,-1).

Answer 1

Suppose f(x) is a function that satisfies both of the following.

1. f(x) is continuous on the closed interval [a,b].
2. f(x) is differentiable on the open interval (a,b).

Then there is a number c such that a < c < b and

f′(c)=[f(b)−f(a)]/(b−a)

(1) f(x)=3x²+1

f'(c) = [f(b)-f(a)]/(b-a)

(a,b)= (0,2)

meaning c is differentiable on the interval (0,2).

f(a) =f(0)=3(0)² + 1 = 1

f(b) =f(2)=3(2)² + 1 = 13

The mean value is:

f'(c) = (13-1)/(2-0) =12/2 = 6

The derivative of f(x) is:

f'(x) =6x

Therefore

f'(c)= 6c = 6

c=1 (It verifies that c in the interval (0,2))

(2) f(x) =x³ -1 over (-2,4)

(-2,4) = (a,b)

f(a)= f(-2) = (-2)³ - 1 = -9

f(b)= f(4) = (4)³ - 1 = 63

f'(c) = (63 - (-9))/(4-(-2)) = (63+9)/(4+2) = 72/6 = 12

For the derivative of f(x):

f'(x) = 3x²

Therefore:

f'(c) = 3c² = 12

c² = 4

c = ± 2?

-2 is extraneous solution for c because it does not belong in the interval (a,b) according to Mean Value Theorem. Therefore:

c = 2

(3) Given f(x) = √(x-1) + 2, over (2,10)

f(a) = f(2)= √(2-1) + 2 = 1+2 = 3

f(b) = f(10)= √(10-1) + 2 = 3+2 = 5

Therefore:

f'(c) = (5-3)/(10-2) = 2/8 = 1/4

The derivative of f(x) is:

f'(x) = 1/(2√(x-1))

Therefore:

f'(c) = 1/(2√(c-1)) = 1/4

2√(c-1) = 4

√(c-1) = 2

c-1 = 4

c = 5

(4) f(x) = √(x) - 2 over (0,1/2).

f(a) = f(0) = √(0) - 2 = -2

f(b) = f(1/2) = √(1/2) - 2 = √(2)/2 - 2

f'(c) =(√(2)/2 - 2-(-2))/(1/2-0)

f'(c) =(√(2)/2 - 2-(-2))/(1/2-0) = √2

The derivative of f(x) is

f'(x)= 1/(2√x)

Therefore:

f'(c) =1/(2√c) = √2

1/(2√2) = √c

1/(4•2) = c

c=1/8

5. Given f(x) = 2x/3, over (-5,-1).

f(a) = f(-5)=2(-5)/3 = -10/3

f(b) = f(-1) = 2(-1)/3 = -2/3

f'(c) = (-2/3 + 10/3)/(-1+5)

f'(c) = 2/3

and the derivative of f(x) is

f'(x) =2/3

Variables c and x are eliminated when you get the derivative because the function itself is linear therefore the answer is not one number but all the numbers in the interval.

c ∈ (-5,-1)

Answer 2

To find the mean value of any continuous function on a closed interval:

f_avg = ( ∫_a^b f(x)dx) / (b-a) . This formula calculates the height of a rectangle of width (b-a) that would have the same area as the area under f(x). The MVT for integrals also guarantees that the function f attains this y-value, f_avg, at some time on the interval (a,b). Thus, there is a c with a < c < b such that f(c) = f_avg.

1) f_avg = ( ∫₀² 3x² + 1 dx) / (2 - 0) = [x³ + x]₀² / 2 = 10/2 = 5 3x² + 1 = 5 ; x² = 4/3 ; x = 2/√3 ...