Question

Does there exist a function \(f\) such that \(f\left( 0 \right) = - 1,{\rm{ }}f\left( 2 \right) = 4,{\rm{ and }}f'\left( x \right) \le 2\), for all \(x\)?

Short answer

No such function exists.

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 for a given condition.

Let any function \(f\left( x \right)\)exist that follows givenconditions:

\(f\left( 0 \right) = - 1,{\rm{ }}f\left( 2 \right) = 4,{\rm{ and }}f'\left( x \right) \le 2\)

Then, according to the mean value theorem, we have:

\(f\left( 2 \right) - f\left( 0 \right) = f'\left( c \right)\left( {2 - 0} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {{\rm{for }}c \in \left( {0,2} \right)} \right\}\)

Solving this, we get:

\(\begin{aligned}{c}f'\left( c \right) &= \frac{{4 - \left( { - 1} \right)}}{2}\\ &= \frac{5}{2}\end{aligned}\)

But, according to the question, \(f'\left( x \right) \le 2,\,\,\,\,\,\,\forall x \in \mathbb{R}\).

This contradicts the condition.

Hence, there is no such function going to exist.