Question

What are the local extrema of `f(x) = cos(x)/x^2+2x^3-x`?

Answer 1

`xapprox 0.8433440445`

Explanation:

we get

`f'(x)=-sin(x)/x^2-2cos(x)/x^3+6x^2+1`
so `f'(x)=0`
if
`6x^5-x^3-xsin(x)-2cos(x)=0`
using a numerical method we get
`xapprox 0.8433440445`