Question

Find the critical points, local max and min, and absolute min and max values, if they exists.

f(x)=x^2/x^2-1 on interval [-4,4]

Answer 1

Okay, let's begin with the domain of this function, or more simple.
What values cannot x take?

Answer 2

My only problem is
for finding the local extrema.
I got a derivative of
\[\frac{ -2x }{ (x^2-1)^2 }\]
and my critical points are
when x is +1 or -1.
So the intervals for testing
points are from
-infinity to -4, -4 to -1,
-1 to 1, and from 1 to infinity.

Answer 3

Now when I plug in my test points at the derivative, I get pos from -infinity to -4 and from -4 to -1, neg from -1 to 1, neg from 1 to 4 and neg from 4 to infinity, so I know by the rule of local extrema that theres local max at 0, but I'm not sure why there would be local min at -4 and 4 according to wolframalpha's answer.

Answer 4

I'll start from the beginning:
having the folowwing function:
\[f:f(x)=\frac{ x ^{2} }{ x ^{2}-1 }\]
The first step is to analyze the domain:
\[D(f)=\mathbb{R} - \left\{ -1,1 \right\}\]
By that I mean "The domain of this function are all the real numbers except '-1' and '1'".
This is a very important thing to do because now I have asympthotes parallel to the Y-axis.
It would look like this:

That means that no extrema can exist there.
Let's analyze the limits as we approach them:
\[\lim_{x \rightarrow -1^{+}}\frac{ x ^{2} }{ x ^{2}-1 }=- \infty \]
\[\lim_{x \rightarrow -1^{-}} \frac{ x ^{2} }{ x ^{2}-1 }=+ \infty \]
\[\lim_{x \rightarrow 1^{+}}\frac{ x ^{2} }{ x ^{2}-1 }=- \infty\]
\[\lim_{x \rightarrow 1^{+}} \frac{ x ^{2} }{ x ^{2}-1 }= + \infty\]
So that tells us that the extremas will actually never ocurr in x= -1 and x=1 because the function is not continous there.

Answer 5

Let's now derive this function:
\[f:f(x)=\frac{ x ^{2} }{ x ^{2}-1 }\]
Then, deriving:
\[f'(x)=\frac{ 2x(x ^{2}-1)-(2x)(x ^{2}) }{ (x ^{2}-1)^{2} }\]
\[f'(x)=\frac{ 2x ^{3}-2x-2x ^{3} }{ (x ^{2}-1)^{2} }=\frac{ -2x }{ (x ^{2}-1)^{2} }\]
\[f'(x)=\frac{ -2x }{ (x ^{2}-1)^{2} }\]
Let's now study the sign of these, let's start with the numerator:

Since I'm only interested in the region from -4 to 4, I'l only study that little part and the rest I'll discard.
Let's do the denominator now:

Let's now do, the total sign:

So we can conclude it has a maximum on x=0.
The critical points are found by doing the exact same process but with the 2nd derivative.