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I am asked to find How many critical points does the function $g(x) = |x^2 − 4|$ have? I know that the result is $3$ but I can only find $2$. What I do, is to equal the equation to $0$, so $x^2-4=0$ and then I factorize so $(x-2)(x+2)$, so I would say that the only critical points happen is $x=-2$ and $x=2$, so two critical points? Where is the third one? How do I find it?

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  • $\begingroup$ Definitions of critical point differ. Usually it is the places where the derivative is $0$, plus the points where the derivative doesn't exist. For critical point at $x=0$ because the derivative is $0$ there, and at $x=\pm 2$ because of the non-differentiability there. $\endgroup$ – André Nicolas Feb 18 '13 at 5:23
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The critical points of $g$ aren’t the values of $x$ at which $g(x)=0$: they’re the values of $x$ at which $g'(x)$ is $0$ or undefined. If you look at the graph of $y=|x^2-4|$, you’ll see that it has a horizontal tangent at $x=0,y=4$ and sharp points at $x=\pm2,y=0$; those three points are the critical points.

To do it analytically, get rid of the absolute value: $x^2-4\ge0$ if and only if $-2\le x\le 2$, so

$$g(x)=\begin{cases} x^2-4,&x\le-2\text{ or }x\ge 2\\ 4-x^2,&-2\le x\le 2\;. \end{cases}$$

Now

$$g'(x)=\begin{cases} 2x,&x<-2\text{ or }x>2\\ -2x,&-2<x<2\;. \end{cases}$$

What happens where the two cases meet, at $x=-2$ and $x=2$? The derivatives on the two sides disagree, and $g'(-2)$ and $g'(2)$ are therefore undefined. That gives you two critical points, and the third is $x=0$, found by solving $g'(x)=0$: the first case gives no solution $-$ you can’t have $2x=0$ if $x<-2$ or $x>2$ $-$ and the second gives the solution $x=0$.

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Critical points are not where the function is $0.$ You want to find the points where the derivative is $0.$ Unfortunately, your function does not happen to be differentiable at $2$ or $-2,$ so you should only get one critical point (at $0$).

Edit: Brian Scott is correct - critical points also include when the derivative is undefined, so $\pm 2$ do count.

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