Question

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

Short answer

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

Step-by-step solution

Given data

The values are \(f(1) = 10\) and \({f^\prime }(x) \ge 2\) for \(1 \le x \le 4\).

Concept of Mean value theorem

“Let\(f\)be a function that satisfies the following hypothesis:

1.\(f\)is continuous on the closed interval\(\left( {a\;,{\rm{ }}b} \right)\).

2.\(f\)is differentiable on the open interval\(\left( {a,b} \right)\).

Then, there is a number\(c\)in\(\left( {a\;,{\rm{ }}b} \right)\)such that\({f^\prime }(c) = \frac{{f(b) - f(a)}}{{b - a}}\).

Or, equivalently,\(f(b) - f(a) = {f^\prime }(c)(b - a)\).”

Calculation to find the smallest possible value of \(f(4)\)

From the given conditions, it is observed that the values of \(a = 1\) and \(b = 4,f(a)\) is represented by \(f(1) = 10\), \({f^\prime }(c)\) is given by \({f^\prime }(x) \ge 2\) and to find possible \(f(4)\) which is represented by \(f(b)\).

Substitute the given values in \({f^\prime }(c) = \frac{{f(b) - f(a)}}{{b - a}}\).

\(\begin{aligned}{c}{f^\prime }(c) &= \frac{{f(4) - f(1)}}{{4 - 1}}\\{f^\prime }(c) &= \frac{{f(4) - 10}}{3}\\f(4) - 10 &= {f^\prime }(c)(3)\\f(4) &= 3{f^\prime }(c) + 10\end{aligned}\)

Simplify further to check the smallest possible value of \(f(4)\) and substitute \({f^\prime }(x) \ge 2\) for \({f^\prime }(c)\).

\(\begin{aligned}{c}f(4) \ge 3(2) + 10\\f(4) \ge 6 &+ 10\\f(4) \ge 16\end{aligned}\)

Thus, the smallest value of \(f(4)\) is \(16\).