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Let $f(x)$ be continuous on $(a,b)$ and differentiable except at a point $c\in(a,b)$.

(note that $f$ may well be differentiable at $x=c$, it's just that we don't take this into our assumption)

If $f'(x)$ changes from positive to negative at $c$, then $c$ is a local maximum.

If $f'(x)$ changes from negative to positive at $c$, then $c$ is a local minimum.

If $f'(x)$ is of the same sign on both sides of $c$, then $c$ is not an extremum.

This essential fact underlies the method described in this HowDo and other tests using higher derivatives.

We want to find the extrema of a given function, which we assume to have a finite number of isolated critical and singular points. This is often this case of a function encountered during exam.

If the function is defined piecewise, then we may need to find the left and right derivative at certain points.

To find the critical points, we set $f'(x) = 0$ and find the solutions.

To find the singular points, we find all the points where $f'(x)$ does not exist.

Mark all the CPs and SPs on the real axis and divide the domain of definition into intervals.

There's a theorem (Darboux's theorem) in Calculus that tells us the sign does not change in each of the intervals obtained in Step 4.

So, as a shortcut, you could find the sign of $f'$ at one point for each interval, or you can also calculate the values of the function at the endpoints, which are needed for the maximal or minimal values anyways.

We classify each CP and SP according to the sign of $f'(x)$ on the left and on the right. The following classification is justified by the fact we mentioned at the beginning.

Consider the function $f(x) = - x^2$. We know $f'(x) = -2x$.

So the function has one critical point and no singular point.

We know $f'>0$ when $x<0$, and $f'<0$ when $x>0$, so this is a case of "Left: +, Right: -", and the point is a local maximum.

This is the function $f(x) = |x|^{1/2}$. The function is not differentiable at $x=0$, which means that $x=0$ is a singular point. This is a local minimum.

The function is $f(x) = x^3$. The derivative $f'(x) = x^2$ is positive on the left and on the right of $x=0$, which is a critical point.

Here we consider a more complex case. We want to find all the local extrema of the function

$$f(x) = 3(x+1)^{\frac{2}{3}}-|x|. $$

This provides an instructive example involving many cases discussed before, and we will cover this in the next four cards!

First of all, we need to find all the CPs, which requires us to differentiate the functions.

We find

$$g'(x)=2(x+1)^{-1/3},$$

$$h'(x)= \begin{cases}-1, & \text { if } x<0 \\ +1, & \text { if } x > 0\end{cases}$$

Therefore, by solving $f'(x) = g'(x)-h'(x) = 0$, we get $x=-9$ and $x=7$. So the critical points are $x = -9$ and $x = 7$.

Next step is to find the SPs.

We should notice that the function is a difference of $g(x) = 3|x+1|^{\frac{2}{3}}$ and $h(x) = |x|$. The first function has an SP at $x=-1$ and at no other point; the latter has an SP at $x=0$ and at no other point. So the difference has two SPs, at $x=-1$ and $x=0$.

To summarize, the CPs are $x=-9,\ x=7$, and the SPs are $x=-1,\ x=0$.

The intervals are $(-\infty, -9)$, $(-9,-1)$, $(-1,0)$, and $(0,\infty)$.

We could solve the inequality $f'(x) > 0$ to get the intervals where this is the case, which gives us this diagram above. Alternatively, we could use the tricks mentioned in Step 5.

At last, we may make the following classification:

Here's what the function really looks like!

Note that $x=0$ is a singular point, but not a local extremum.

If the question is to determine the global extrema, you should also consider the boundary of the domain. The procedure is to determine all local maximal value and local minimal value, and compare them against each other and the values at the boundary.

So be careful when the question asks you to find global maxima and minima!