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

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Answer 1 · 2018-06-02T15:54:17

$xapprox 0.8433440445$

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
$xapprox 0.8433440445$

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