What are the minimum and maximum of $f(x) = x^6 - 5x^4 + 5x^2 - 1$?

Question

I am stuck in obtaining the critical points:

$f'(x) = 6x^5 - 20x^3 + 10x$

$f''(x) = 30x^4 - 60x^2 + 10$

For critical points, we have: $f'(x) = 0$

$2x(3x^4 - 10x^2 + 5) = 0$

Now, how should i solve them to get the critical points. And what should the minima and maxima of the given function be at the end?

Answer 1

$2x(3x^4 - 10x^2 + 5) = 0$

$x_1=0$

$3x^4 - 10x^2 + 5 = 0$

Let $y=x^2$ then $3y^2 - 10y + 5 = 0$ and $y_{1,2}=\frac{5\pm\sqrt{10}}{3}$

and

$x_2=-\sqrt{\frac{5-\sqrt{10}}{3}}$, $x_3=\sqrt{\frac{5-\sqrt{10}}{3}}$, $x_4=-\sqrt{\frac{5+\sqrt{10}}{3}}$, $x_5=\sqrt{\frac{5+\sqrt{10}}{3}}$

So the maximum is $\infty$

Function is minimum when $x=-\sqrt{\frac{5+\sqrt{10}}{3}}$ or $x= \sqrt{\frac{5+\sqrt{10}}{3}}$

Answer 2

Note the symmetry of coefficients:

$\displaystyle {f(x) \over x^3} = (x^3-{1\over x^3}) - 5·(x-{1\over x}) = (x-{1\over x})^3 - 2·(x-{1\over x}) $

This suggested we can simplify, letting $u = x^2-1$

$f(x) = g(u) = u^3 - 2u^2 - 2u$

For extremum, just solve $g'(u) = 0$, then, back solve for x