How to prove that the local minimum for $f(x)=\left|x\right|$ occurs at $x=0$?

Question

The first and second derivative tests don't work as $f\left(x\right)$ is not differentiable at $x=0$. So, critical points can't be obtained to check for local maxima or minima or inflection.

Plotting the graph for $f\left(x\right)=\left|x\right|$ does show that $f'\left(x\right)$ or $\frac{df\left(x\right)}{dx}$ goes from decreasing to increasing as $f\left(x\right)$ passes $x=0$. Thus, $x=0$ can be termed as the point of local minimum.

However, can this be done through another method in addition to curve sketching just like the first and second derivative tests?

Answer 1

No, you can still apply the first derivative test when $f(x)$ is not differentiable at $x=0$.

From definition, $0$ is a critical point of the function $f$ even $f$ is not differentiable at $0$.

To apply the first-derivative test, you only need $f$ is continuous at $0$, $f$ is differentiable on $(-\delta, 0)$ and $(0, \delta)$ for some $\delta>0$.

Since $$f^{'}(x) \ge 0 \: \text{ for all } x\in(0, \delta) \implies f(x) \text{ is increasing on } (0, \delta)$$

$$f^{'}(x) \le 0 \: \text{ for all } x\in(-\delta, 0) \implies f(x) \text{ is decreasing on } (-\delta, 0)$$

$f$ has a local minimum at $x=0$.

More than that, since $f(x)=|x| \ge 0$ and $f(0)=0$, $f$ has a global minimum at $x=0$.

Answer 2

You don't need to sketch the graph. According to the definition, $|x|$ is always $\ge0$.

So the minimum will occur when $|x|=0$. In another words, minimum will occur when $x=0$.

Answer 3

First consider the interval $]0,\infty[$, in which $f(x)=x$ and thus $f'(x)=1$. Since $f$ has a derivative on this interval and $f'$ is never equal to 0, no value $x>0$ can be a local minimum (or maximum).

Second consider the interval $]-\infty,0[$, in which $f(x)=-x$ and thus $f'(x)=-1$. Since $f$ has a derivative on this interval and $f'$ is never equal to 0, no value $x<0$ can be a local minimum (or maximum).

There remains only one value to consider, namely $x=0$ where $f(x)=0$, which appears to be a global minimum (since $\forall x \in \mathbb{R}, |x| \geq 0$).

To conclude, the function $f$ has a unique local extremum in $0$, which happens to be a global minimum.