Question

Suppose \(3 \le f'\left( x \right) \le 5\), for all values of \(x\). Show that \(18 \le f\left( 8 \right) - f\left( 2 \right) \le 30\).

Short answer

It is proved that \(18 \le f\left( 8 \right) - f\left( 2 \right) \le 30\) in the interval \(\left( {2,8} \right)\).

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 for a given condition

The givenconditions are:

\(3 \le f'\left( x \right) \le 5\,\,\,\,\,\,\,\,\,\,\,\,\forall x \in \mathbb{R}\)

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

\(f\left( 8 \right) - f\left( 2 \right) = f'\left( c \right)\left( {8 - 2} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {{\rm{for }}c \in \left( {2,8} \right)} \right\}\)

By relating the given condition, we get:

Hence proved: \(18 \le f\left( 8 \right) - f\left( 2 \right) \le 30\) in the interval \(\left( {2,8} \right)\).