So I'm ask to prove using the mean value theorem that $\sin(x) \ge x-\frac{x^3}{6}$

I understand that the mean value theorem works because both sides of the equation equal zero when $x=0$.

To start I set $f(x)=\sin(x)$ and $g(x)=x-\frac{x^3}{6}$ and let $F(x)=f(x)-g(x)$ and try to show that $F'(x)>0$ for all $x>0$.

Therefore, we have $F(x)=\frac{F(x)-F(0)}{x-0}(x-0)$

By the mean value theorem, we know there exists a $y>0$ such that $F'(y)x=(\cos(x)-1+\frac{x^2}{2})x$

If we apply MVT again for some $z>0$, we get $F'(z)x=\frac{F'(y)-0}{x}x=(-\sin(x)+2x)x$

It seems pretty obvious that $F'(z)x>0$ but just for safe measure I'll apply it again: There exists a $c>0$ such that $F'(c)x=(-\cos(x)+2)x$

$\therefore$ since $x>0$ and $2-\cos(x)\le 1$ (by the bounds of cosine), $F'(c)>0$ for $c>0$

$ \therefore F(x)>0\equiv f(x)>g(x)$ if $x>0 $

I have my final tomorrow so any tips to make this more concrete would be appreciated


You have the right idea. Show in succession, using MVT, that for $x>0$, $x>\sin x$, $\cos x>1-x^2/2$, $\sin x> x-x^3/6$.


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