Finding extrema of $f(x) = \frac {\sqrt{x^4 - 4x^2 + 4}}{x^2 -1}$

Question

Find all of the extreme points of the following function and sketch a graph.

$f(x) = \frac {\sqrt{x^4 - 4x^2 + 4}}{x^2 -1}$

So far what I did is simplify the function to get the following: $f(x) = \frac {|(x+\sqrt2)(x - \sqrt2)|}{(x-1)(x+1)}$. From this form it is easy to get both the x and y intercepts and the domain of the function as well as the vertical asymptotes at $x =\pm1$.

How would I find the extrema of this function, and give proper reason for them being the extrema and also the type of extreme that they are? (i.e. local, global minimum or maximum, not being an inflection point etc.)

Answer 1

Hint: By the quotient and the chain rule we get $$f'(x)=2\,{\frac { \left( {x}^{2}-2 \right) x}{\sqrt { \left( {x}^{2}-2 \right) ^{2}} \left( x-1 \right) ^{2} \left( x+1 \right) ^{2}}} $$ and solve the equation $$f'(x)=0$$ for $$x$$. The simplified second derivative is given by $$f''(x)-2\,{\frac { \left( 3\,{x}^{2}+1 \right) \left( {x}^{2}-2 \right) ^{3 }}{ \left( \left( {x}^{2}-2 \right) ^{2} \right) ^{3/2} \left( x-1 \right) ^{3} \left( x+1 \right) ^{3}}} $$

Answer 2

For the sake of simplification, let us use the change of variable $t=x^2$. Then

$$\frac{|t-2|}{t-1}=\pm\frac{t-2}{t-1}=\pm\left(1-\frac1{t-1}\right)$$ has no stationary point. Anyway, close to $t=2$, $|t-2|$ is forming a non-differentiable local minimum (while $t-1$ is roughly the constant $1$).

To this, you need to add $x=0$, by the chain rule.

To summarize,

$$x=-\sqrt2,0,\sqrt2.$$