Question

The value of c for which the mean value of the theorem hols for the function f(x) = 2x - x 2 on the interval [0,1] is:

• A. 0
• B. $\frac{1}{4}$
• C. $\frac{1}{2}$
• D. $\frac{1}{3}$

Answer 1

Answer: (C)

$\frac{1}{2}$

Answer 2

The Mean Value Theorem states that if f is continuous on [a, b] and differentiable on (a, b) then there exists some number c in (a, b) such that:
$[ f'(c) = \frac{f(b) - f(a)}{b - a} ]$

Given:
$[ f(x) = 2x - x^2 ]$ on the interval [0,1].

First, find
$( f'(x) ): [ f'(x) = 2 - 2x ]$

Next, evaluate the function at the endpoints of the interval:
$[ f(1) = 2(1) - (1)^2 = 1 ] [ f(0) = 0 ]$

Substitute the given values into the formula from the Mean Value Theorem:
$[ 2 - 2c = \frac{1 - 0}{1 - 0} = 1 ]$

Solving for $( c ): [ 2 - 2c = 1 ] [ -2c = -1 ] [ c = \frac{1}{2} ]$

So, the correct answer is: C. $( c = \frac{1}{2} )$