Prove that if $|g'(x)|\leq M|x-x_0|^n$ with $|x-x_0|<\delta$ then $|g(x)|\leq \frac{M|x-x_0|^{n+1}}{n+1}$

Question

I really don't know how to proceed. There is similar question here but in that case, they prove that $|g(x)-g(a)|\leq \frac{M|x-x_0|^{n+1}}{n+1}$ instead of just $|g(x)|$ so I got really confused.

I hope that yoy could help me.

score 0 · Answer 1 · answered Jun 04 '18 at 05:08

0

The assertion is not correct. Take $g(x)=x^2+7, x_0=0$ and $n=1$.

answered Jun 04 '18 at 05:08

Fred

77,394

Prove that if $|g'(x)|\leq M|x-x_0|^n$ with $|x-x_0|<\delta$ then $|g(x)|\leq \frac{M|x-x_0|^{n+1}}{n+1}$

1 Answers1