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

The Correct Option is C

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. \(( c = \frac{1}{2} )\)