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?
Best Answer
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 $x=0$.