Question

Let \(f\left( x \right) = 2 - \left| {2x - 1} \right|\). Show that there is no value of \(c\) such that \(f\left( 3 \right) - f\left( 0 \right) = f'\left( c \right)\left( {3 - 0} \right)\). Why does this not contradict the Mean Value Theorem?

Short answer

There is no number \(c\)for which the relation \(f\left( 3 \right) - f\left( 0 \right) = f'\left( c \right)\left( {3 - 0} \right)\) is valid.

And the function does not contradict the Mean Value Theorem as \(f\left( x \right)\) is not differentiable at \(x = \frac{1}{2}\) which is in domain \(\left( {0,3} \right)\).

Step-by-step solution

Mean Value Theorem

The Mean Value Theorem is applicable for any function \(f\left( x \right)\) if and only if

\(\begin{array}{l}f\left( x \right) \to {\rm{continuous within }}\left( {a,b} \right)\\f\left( x \right) \to {\rm{differentiable within }}\left( {a,b} \right)\\{\rm{then, for }}c \in \left( {a,b} \right) \to f'\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}}\end{array}\).

Solving the given function:

The given function is:

\(\begin{aligned}{c}f\left( x \right) &= 2 - \left| {2x - 1} \right|\\ &= \left\{ \begin{array}{l}2 - \left( {2x - 1} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{if }}2x - 1 \ge 0\\2 - \left( { - \left( {2x - 1} \right)} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{if }}2x - 1 < 0\end{array} \right.\\ &= \left\{ \begin{array}{l}3 - 2x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{if }}x \ge \frac{1}{2}\\1 + 2x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{if }}x < \frac{1}{2}\end{array} \right.\end{aligned}\)

On differentiating, we get:

\(f'\left( x \right) = \left\{ \begin{array}{l} - 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{if }}x > \frac{1}{2}\\2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{if }}x < \frac{1}{2}\end{array} \right.\)

Solving as per the condition of the mean value theorem:

On solving, we have:

\(\begin{aligned}{c}f\left( 3 \right) - f\left( 0 \right) &= f'\left( c \right)\left( {3 - 0} \right)\\ - 3 - 1 &= f'\left( c \right)\left( 3 \right)\\f'\left( c \right) &= - \frac{4}{3}\end{aligned}\)

Since, \(f'\left( c \right) \ne \pm 2\).

Hence, there is no such number \(c\)exists that satisfies the theorem.

Also, the function does not contradict the Mean Value Theorem because \(f\left( x \right)\) is not differentiable at \(x = \frac{1}{2}\) which is in domain \(\left( {0,3} \right)\).