Question

A function f(x) = 1 - x² + x³ is defined in the closed interval [-1, 1]. The value of x, in the open interval (-1, 1) for which the mean value theorem is satisfied, is
1. -1/2
2. -1/3
3. 1/3
4. 1/2

Answer 1

Correct Answer - Option 2 : -1/3

By Lagrange’s mean value theorem we have

\(f'(c) = \frac{{f(b) - f(a)}}{{b - a}}\)

\(\begin{array}{l} {{\rm{f}}'}\left( {\rm{x}} \right) = \frac{{{\rm{f}}\left( 1 \right) - {\rm{f}}\left( { - 1} \right)}}{{1 - \left( { - 1} \right)}}\\ \end{array}\)

f(1) = 1 - 1 + 1 = 1

f(-1) = 1 - 1 - 1 = -1

\(f^1(x) = \frac{2}{2} = 1{\rm{\;}}\\ \)

0 - 2x + 3x² = 1 ⇒ 3x² - 2x - 1 = 0

\(\therefore {\rm{\;x\;}} = {\rm{\;}}1{\rm{\;and\;}} - \frac{1}{3}\\ \therefore {\rm{\;x}} = - \frac{1}{3}{\rm{\;only\;lies\;in\;}}\left( { - 1,{\rm{\;}}1} \right) \)