Let $f(x)=x+x^{100}+x^{1000}$. Prove that there exists an interval $(-\epsilon ,\epsilon )$ where $f(x)$ is increasing.
(My attempt) $f'(x)=1+100x^{99}+1000x^{999}\geq 0 $
So, if $x\geq 0$ then $f(x)$ is increasing. Next I want to show that $f'(x)=1+100x^{99}+1000x^{999}\geq 0 $ for every $y$ in $(0,-\epsilon) $ for some $\epsilon > 0$
I think Intermediate Value Theorem can be applied here, but I don't know how to proceed.
Notice that $f'(x) = 1+100x^{99} + 1000x^{999}$ is continuous. Since $f'(0)=1$, there exists a neighborhood of $0$ such that $f'(0)>1/2$
Note that $f''(x)=9900x^{98}+999000x^{998}\geq 0$ which implies that $f'$ is increasing. Moreover $$f'(-1/2)=1-100/2^{99}-1000/2^{999}>1-1/2-1/2=0.$$ Hence we can take $\epsilon=1/2$.
P.S. If you are not interested in an explicit value of $\epsilon$ then use the fact that $f'$ is continuous at $0$ and $f'(0)=1>0$.
Let $\max\{|x|,|y|\}=:h$. Then $$f(y)-f(x)=(y-x)\left(1+\sum_{k=0}^{99}y^{99-k}x^k +\sum_{k=0}^{999}y^{999-k}x^k\right)\ .$$ The two sums $\Sigma_i$ on the right hand side can be estimated by $$|\Sigma_1|+|\Sigma_2|\leq 100 h^{99}+1000 h^{999}<1$$ when $h$ is sufficiently small.
