$\displaystyle f(x)=x^{4}+(sinx)e^{-x^{2}}$

prove that if $\displaystyle x_{0}$ is the minimum point

then $\displaystyle f(x_{0})<0$

hint: look at f(0) and f'(0).

how i tried to solve it.

f(0)=0 f'(0)=1

by limit derivative definition $\displaystyle f'(0)=lim_{x->0}\frac{f(0+x)-f(0)}{x-0}=lim_{x->0}\frac{f(x)}{x}=1$

so by limit definition for $\displaystyle -\delta<x<\delta$

$\displaystyle 1-\epsilon<\frac{f(x)}{x}<1+\epsilon$

what now?