Question

Does there exist a function \(f\)such that \(f(0) = - 1\;,\;f(2) = 4\) and \({f^\prime }(x) \le 2\) for all x?

Short answer

There is no such function exist.

Step-by-step solution

Given data

The given conditions are \(f(0) = - 1\;,\;f(2) = 4\) and \({f^\prime }(x) \le 2\) for all \(x\).

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 check the function exists or not

Suppose that such function \(f\) exists, then by the Mean value Theorem mentioned above, there exist a number \(c\) in the interval \((0,2)\) such that, \({f^\prime }(c) = \frac{{f(2) - f(0)}}{{2 - 0}}\).

Substitute the respective values that given in the conditions.

\(\begin{aligned}{c}{f^\prime }(c) &= \frac{{f(2) - f(0)}}{{2 - 0}}\\{f^\prime }(c) &= \frac{{4 - ( - 1)}}{2}\\{f^\prime }(c) &= \frac{{4 + 1}}{2}\\{f^\prime }(c) &= \frac{5}{2}\end{aligned}\)

From the above equation, it is obtained that \({f^\prime }(c) = \frac{5}{2}\) and it is given that, \({f^\prime }(x) \le 2\).

This implies that, \(\frac{5}{2} = {f^\prime }(x) \le 2\).

This is not possible.

This contradicts our assumption that there exists such function.

Thus, it can be concluded that there is no such function exist.