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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?

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    $\begingroup$ Just check the definition of a minimum: For all $x\in \Bbb R$ is $f(x) = |x| \ge 0 = f(0)$. $\endgroup$
    – Martin R
    Jul 28, 2023 at 12:08
  • $\begingroup$ You can use $|x| \ge 0$ $\endgroup$
    – Tim
    Jul 28, 2023 at 12:08
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    $\begingroup$ The critical point test still works when $f'$ is undefined as long as $f$ is still continuous there. Most definitions of critical point allow for $f'$ undefined precisely because of this. $\endgroup$
    – Randall
    Jul 28, 2023 at 12:23
  • $\begingroup$ Something overkill here, though more general as well, would be the sub gradient and convexity. $\endgroup$
    – Andrew
    Jul 28, 2023 at 12:36
  • $\begingroup$ For this problem, using derivatives at all is the hard way! $\endgroup$
    – aschepler
    Jul 28, 2023 at 16:26

3 Answers 3

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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 $x=0$.

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    $\begingroup$ Ohh yes, makes sense. Critical points can also occur at values of x where f '(x) is undefined. So, the first derivative test can still be applied to prove x=0 is a point of extremum, specifically, point of relative/local minimum which in this case is also the absolute or global minimum. $\endgroup$
    – user3.14
    Jul 28, 2023 at 13:20
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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 $x=0$.

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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.

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  • $\begingroup$ Yes, got it. Thank you! $\endgroup$
    – user3.14
    Jul 28, 2023 at 14:16

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