If $x_{0}\in(a,b)$, $f$ is differentiable at $x_{0}$, and $f$ attains either a local maximum or local minimum at $x_{0}$, then $f'(x_{0}) = 0$.

Question

Let $a < b$ be real numbers, and let $f:(a,b)\rightarrow\textbf{R}$ be a function. If $x_{0}\in(a,b)$, $f$ is differentiable at $x_{0}$, and $f$ attains either a local maximum or local minimum at $x_{0}$, then $f'(x_{0}) = 0$.

MY ATTEMPT

Let us consider the local minimum first. Thus there exists $\delta > 0$ such that $f(x)\geq f(x_{0})$ for $|x - x_{0}| < \delta$.

If $x_{0} < x < x_{0} + \delta$, then we have that \begin{align*} f(x) \geq f(x_{0}) \Longrightarrow f(x) - f(x_{0}) \geq 0 \Longrightarrow \frac{f(x) - f(x_{0})}{x - x_{0}} \geq 0 \Longrightarrow f'(x^{+}_{0}) \geq 0 \end{align*}

If $x_{0} - \delta < x < x_{0}$, then we have that \begin{align*} f(x) \geq f(x_{0}) \Longrightarrow f(x) - f(x_{0}) \geq 0 \Longrightarrow \frac{f(x) - f(x_{0})}{x - x_{0}}\leq 0 \Longrightarrow f'(x^{-}_{0})\leq 0 \end{align*}

Since $f$ is differentiable at $x_{0}$, we conclude that $f'(x^{+}_{0}) = f'(x^{-}_{0})$, which is only possible if $f'(x_{0}) = 0$, and we are done.

Similar reasoning applies to the case when $x_{0}$ is a local maximum.

Could someone verify if I am reasoning correctly?

Answer 1

Yes, what you did is right. By the way, to treat the local maximum case, here's a cute trick if you haven't already seen it: if $f$ has a local maximum at $x_0$, then $-f$ has a local minimum at $x_0$. So, $(-f)'(x_0) = -f'(x_0) = 0$. Hence, $f'(x_0) = 0$. So, this saves you the trouble of going through the entire proof again.

Answer 2

This looks good to me!

Another way of proving this (which I had done during my undergrad was by contradiction)

Suppose $f$ has a local minimum(or maximum) at $x_0$. Now, if possible let us suppose that $f'(x_0)\not= 0$.

Then going along the same line of argument as yours , you shall be able to argue that it(i.e. $f'(x_0)\not= 0$) contradicts the fact that $x_0$ is local minimum (or maximum).

Let me show it here for you.

Let us consider the local minimum first. Thus there exists $δ>0$ such that $f(x)≥f(x_0)$ for $x\in(x_0-\delta,x_0+\delta)$, $\delta>0$.

If $f'(x_0)> 0$. Then $\exists \delta_1 \in (0,\delta)$ such that $$\frac{f(x) - f(x_{0})}{x - x_{0}} > 0, \forall x\in(x_0-\delta_1,x_0)$$ $$\implies f(x) - f(x_{0}) < 0, \forall x\in(x_0-\delta_1,x_0)$$ $\therefore f(x)<f(x_{0}), \forall x\in(x_0-\delta_1,x_0)\subset(x_0-\delta,x_0+\delta) $ ,this contradicts the fact that $x_0$ is local minimum.

So $f'(x_0)\not> 0$

Similarly you can show that $f'(x_0)\not< 0$. And hence the result.