# What are the global and local extrema of f(x)=2x^7-2x^5  ?

We rewrite f as

$f \left(x\right) = 2 {x}^{7} \cdot \left(1 - \frac{1}{x} ^ 2\right)$

but ${\lim}_{x \to \infty} f \left(x\right) = \infty$ hence there is no global extrema.

For the local extrema we find the points where $\frac{\mathrm{df}}{\mathrm{dx}} = 0$

$f ' \left(x\right) = 0 \implies 14 {x}^{6} - 10 {x}^{4} = 0 \implies 2 \cdot {x}^{4} \cdot \left(7 \cdot {x}^{2} - 5\right) = 0 \implies {x}_{1} = \sqrt{\frac{5}{7}} \mathmr{and} {x}_{2} = - \sqrt{\frac{5}{7}}$

Hence we have that

local maximum at $x = - \sqrt{\frac{5}{7}}$ is $f \left(- \sqrt{\frac{5}{7}}\right) = \frac{100}{343} \cdot \sqrt{\frac{5}{7}}$

and

local minimum at $x = \sqrt{\frac{5}{7}}$ is $f \left(\sqrt{\frac{5}{7}}\right) = - \frac{100}{343} \cdot \sqrt{\frac{5}{7}}$