Question

Prove Fermat's theorem for the case in which \(f\) has a local minimum at \(c\).

Short answer

It is conclude that\(\alpha = 0\)that is\({f^\prime }(c) = 0\).

The Fermat's theorem proved.

Step-by-step solution

Given information

Key idea is that\(f\)has a local minimum at\(c\), then\(f(x) \ge f(c)\)for any\(x\)neighbourhood of\(c\).

Then, \(f(x) - f(c) \ge 0\).

Concept of Fermat's Theorem:

Fermat's Theorem is if\(f\)has a local maximum or minimum at the point\(c\), and if\({f^\prime }(c)\)exists, then\({f^\prime }(c) = 0\).

Check the conditions of Fermat’s Theorem

Key idea is that\(f\)has a local minimum at\(c\), then\(f(x) \ge f(c)\)for any\(x\)neighbourhood of\(c\).

Then,\(f(x) - f(c) \ge 0\).

To prove the Fermat's theorem it is enough to prove\({f^\prime }(c)\)exists.

By the definition,\({f^\prime }(c) = \mathop {\lim }\limits_{x \to c} \frac{{f(x) - f(c)}}{{x - c}}\).

To prove\({f^\prime }(c)\)exists, it is enough to show Left hand limit should be equal to Right hand limit.

Let\({f^\prime }(c) = \alpha \)where\(\alpha \)is constant.

Consider the Left hand limit.

\(\begin{array}{c}\alpha = \mathop {\lim }\limits_{x \to {c^ - }} \frac{{f(x) - f(c)}}{{x - c}}\\\alpha = \frac{{\mathop {\lim }\limits_{x \to {c^ - }} (f(x) - f(c))}}{{\mathop {\lim }\limits_{x \to {c^ - }} (x - c)}}\\\alpha \mathop {\lim }\limits_{x \to {c^ - }} (x - c) = \mathop {\lim }\limits_{x \to {c^ - }} (f(x) - f(c))\end{array}\)

From the key idea mentioned above,\(f(x) - f(c) \ge 0\).

Therefore, Right hand side\(\mathop {\lim }\limits_{x \to {c^ - }} (f(x) - f(c)) \ge 0\)and for the negative values of\(x\mathop {\lim }\limits_{x \to {c^ - }} (x - c) \le 0\).

Therefore,\(\alpha \le 0\). …... (1)

Consider the Right hand limit.

\(\begin{array}{c}\alpha = \mathop {\lim }\limits_{x \to {c^ + }} \frac{{f(x) - f(c)}}{{x - c}}\\\alpha = \frac{{\mathop {\lim }\limits_{x \to {c^ + }} (f(x) - f(c))}}{{\mathop {\lim }\limits_{x \to {c^ + }} (x - c)}}\\\alpha \mathop {\lim }\limits_{x \to {c^ + }} (x - c) = \mathop {\lim }\limits_{x \to {c^ + }} (f(x) - f(c))\end{array}\)

From the key idea,\(f(x) - f(c) \ge 0\).

Therefore, Right hand side\(\mathop {\lim }\limits_{x \to {c^ + }} (f(x) - f(c)) \ge 0\)and for the negative values of\(x,\;\mathop {\lim }\limits_{x \to {c^ + }} (x - c) \ge 0\).

Therefore,\(\alpha \ge 0\). …... (2)

From the equation (1) and (2), it is conclude that\(\alpha = 0\)that is\({f^\prime }(c) = 0\).

Hence, the Fermat's theorem proved.