Question

If \(f\left( 1 \right) = 10{\rm{ and }}f'\left( x \right) \ge 2{\rm{ for }}1 \le x \le 4\), how small can \(f\left( 4 \right)\) possibly be?

Short answer

The smallest value of \(f\left( 4 \right)\) is 16.

Step-by-step solution

Mean Value Theorem

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

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

Solve for the given condition

The givenconditions are:

\(f\left( 1 \right) = 10{\rm{ and }}f'\left( x \right) \ge 2{\rm{ for }}1 \le x \le 4\)

Then, according tomean value Theorem, we have:

\(f\left( 4 \right) - f\left( 1 \right) = f'\left( c \right)\left( {4 - 1} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {{\rm{for }}c \in \left( {1,4} \right)} \right\}\)

By putting the given value, we get:

\(\begin{aligned}{c}f\left( 4 \right) - 10 &= 3f'\left( c \right)\\f\left( 4 \right) &\ge 10 + 3\left( 2 \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {f'\left( x \right) \ge 2} \right\}\\f\left( 4 \right) &\ge 16\end{aligned}\)

Hence, the smallest value of \(f\left( 4 \right)\) is 16.