<s>[INST] Find where $f(x)=\frac{x^2+4}{x^2-4}$ is concave upwards and concave downwards Find where $f(x)=\frac{x^2+4}{x^2-4}$ is concave upwards and concave downwards

Solution: We are going to take the second derivative, find the critical points, and then test each region. Lets get to the second derivative by doing the quotient rule twice:
$f'(x) = \frac{(x^2-4)\frac{d}{dx}(x^2+4)-(x^2+4)\frac{d}{dx}(x^2-4)}{(x^2-4)^2}$
$=\frac{(x^2-4)(2x)-(x^2+4)(2x)}{(x^2-4)^2}$
$=\frac{-16x}{(x^2-4)^2}$
Okay, theres the first derivative. Let's do it again!!
$f''(x)=\frac{(x^2-4)^2 \frac{d}{dx}(-16x) - (-16x)\frac{d}{dx}(x^2-4)^2}{(x^2-4)^4}$
Note that we have to the chain rule to evluate $\frac{d}{dx}(x^2-4)^2$.
$\rightarrow f''(x)=\frac{(x^2-4)^2 (-16) - (-16x)(2(x^2-4)\frac{d}{dx}(x^2-4))}{(x^2-4)^4}$
$\rightarrow =\frac{(x^2-4)^2 (-16) - (-16x)(2(x^2-4)(2x)}{(x^2-4)^4}$
$\rightarrow =\frac{(x^2-4)^2 (-16) + 64x^2(x^2-4)}{(x^2-4)^4}$
Canceling off one of the $(x^2-4)$ from each term in the numerator and the denominator...
$\rightarrow =\frac{(x^2-4) (-16) + 64x^2}{(x^2-4)^3}$
$\rightarrow =\frac{-16x^2+64+64x^2}{(x^2-4)^3}$
$\rightarrow =\frac{64+48x^2}{(x^2-4)^3}$
Few.. okay the hard part is over. Now we need to find the points to chop our number line up from, so we have to solve $f''(x)=0$ and find where $f''(x)$ DNE.
Let's solve where $f''(x)=0$
$f''(x) =\frac{64+48x^2}{(x^2-4)^3}=0$
$\rightarrow 64+48x^2 = 0$
$\rightarrow x= \pm \sqrt{-64/48}$
So $f''(x)=0$ only when $x$ is imaginary... so those values are irrelevant.
On the other hand, $f''(x)$ DNE when $x = \pm 2$, because the denominator will be zero. So these are the values we split our number line up by.
Since $f''(-3)=\frac{496}{125}>0$, we have that $f(x)$ is concave up on $(-\infty,-2)$.
Since $f''(0)=-1<0$, we have that $f(x)$ is concave down on $(-2,2)$.
Since $f''(3)=\frac{496}{125}>0$, we have that $f(x)$ is concave up on $(2,\infty)$.
 [/INST] It's good to keep in mind that the critical points of $f''(x)$ are places where either $f''(x)=0$ or $f''(x)$ is undefined. Hence concavity can change at places where $f''(x)=0$ and at places where $f''(x)$ is undefined.
We were given that
$$f(x)=\frac{x^2+4}{x^2-4},$$
and you correctly calculated that
$$f''(x)=\frac{48x^2+64}{(x^2-4)^3}.$$
Note that there is no $x$ with $f''(x)=0$, but $f''(x)$ is undefined at $-2$ and $2$. So it is enough to check the concavity in the intervals $(-\infty,-2)$, $(-2,2)$, and $(2,\infty)$.
Since $f''(-3)=\frac{496}{125}>0$, we have that $f(x)$ is concave up on $(-\infty,-2)$.
Since $f''(0)=-1<0$, we have that $f(x)$ is concave down on $(-2,2)$.
Since $f''(3)=\frac{496}{125}>0$, we have that $f(x)$ is concave up on $(2,\infty)$.
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