Question

To determine the function\(g(x) = - f(x)\)has a local maximum at\(c\)if\(f\)has a local minimum at\(c\).

Short answer

The function\(g(x) = - f(x)\)has a local maximum value at\(c\).

It is enough to prove that \(g(c) \ge g(x)\) for any \(x\) near to \(c\).

Step-by-step solution

Given function

The function \(f(x)\) has a local minimum at \(c\).

Definition of Local minimum

By the local minimum definition, "if the function\(f(c) \le f(x)\)is local minimum at\(c\), then\(f(c) \le f(x)\)for any\(x\)near to\(c\).”

Check the conditions for the Local Minimum Definition

By the local minimum definition, "if the function\(f(x)\)is local minimum at\(c\), then\(f(c) \le f(x)\)for any\(x\)near to\(c\)."

Thus,\(f(c) \le f(x)\).

Multiply by\(( - 1)\)on both the sides.

\(\begin{aligned}{c} - f(c) \ge - f(x)\\ - f(c) \ge x{\rm{ }}\\g(x) = - f(x)\end{aligned}\)

At a point\(c,\;g(x) = - f(x)\)becomes\(g(c) = - f(c)\).

Therefore, equation (1) becomes,\(g(c) \ge g(x)\).

Therefore,\(g(c) \ge g(x)\)for any\(x\)near to\(c\).

Hence, it is proved that the function \(g(x) = - f(x)\) has a local maximum value at \(c\).