Mean value theorem help?

Question

So I am learning this chapter "mean value theorem", and there is an exercise.

Prove the inequality of $e^{x} \gt x+1$ for $x$ different from zero and the inequality of $2x \arctan x \ge \ln (x^2+1)$... I feel very retarded every time I read this exercise because I don't understand what theorem should I use?

Can you help me just a little and I'll do the rest by myself?

Answer 1

I wouldn't use the mean value theorem there. For the first one:

$e^x>x+1$ since $y=x+1$ is the tangent line to the graph $y=e^x$ in the point $(0,1)$. Since $e^x$ is convex it's graph is above the tangent line.

For the second one I would define $$ f(x) = 2x \arctan x - \log(x^2+1) $$ and study the sign of the derivative $f'(x)$ to prove that $0$ is a global minimum.

addendum Just to see how you get a rigorous proof.

Consider $f(x)=e^x-x-1$. Notice that $f'(x) = e^x-1$. For $x> 0$ we have $f'(x)>0$ hence $f(x)$ is increasing in $[0,+\infty)$. For $x<0$ we have $f'(x)<0$ hence $f(x)$ is decreasing on $(-\infty,0]$. Hence for all $x\neq 0$ we have $f(x) > f(0) = 0$. This means $e^x > x+1$.

For the second one take $g(x) = 2x\arctan x - \log(x^2+1)$. We have $g'(x) = 2\arctan x$ and the proofs is the same as before.

edit So now I realize what you mean by "mean value theorem". In fact to prove that $f'>0$ implies $f$ increasing you use that theorem.

Answer 2

For establishing the inequality $e^x>1+x$, consider two cases: $x>0$ and $x<0$.

For the case $x>0$, apply the Mean Value Theorem to the function $f(x)=e^x$ over the interval $[0,x]$. This gives a $c$ with $0<c<x$ satisfying $$ e^x-e^0 =(x-0)\cdot e^c=xe^c. $$ But $e^0=1$ and $e^c>1$ (strict inequality, since $c>0$). so $$ e^x-1>x; $$ which implies the result for $x>0$.

I'll leave the other case for you...

I'll just give a hint for your second inequality:

For the second inequality, break things up into two cases: $x\ge 0$ and $x<0$.

For the case $x\ge0$, apply the Mean Value Theorem to the function $g(x)=2x\arctan x -\ln(x^2+1)$ over the interval $[0,x]$ and use the fact that $\arctan(c)\ge 0$ for all $c\ge0$. (Note: the second case is easier here, since $g$ is an even function.)