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Can somebody explain to me WHY we can use the MVT to poof inequalities?

First of all what I know:

I know what the MVT says and that is pretty logical. If a function is continuous and differentiable on the interval(a,b) then the MVT says: $\frac{f(b)-f(a)}{(b-a)} = f'(c)$ for AT LEAST one point c in $a<c<b$. That's pretty obvious.

Also if $f'(x)>0 \implies f$ is increasing. If $f'(x) < 0 \implies f$ is decreasing. If $f'(x) = 0 \implies f$ is continous.

But I am missing the part why it actually can be used to prove an inequality.

I tried to prove $\tan(x) > x$ in the interval $0<x<\frac{pi}{2}$

My proof:

$\tan(x)>x \iff \frac{\tan(x)}{x}>1$, now we can add 0 on the numerator and the denominator:

$\frac{\tan(x)-\tan(0)}{(x-0)}>1$

Hence that looks like the difference quotient WE KNOW that there is AT LEAST point $c$ in the interval which then will fulfill:

$f'(c)=\frac{\tan(x)-\tan(0)}{x-0}>1$

Now I took the derivative of $\tan(x)$ which equals $\frac{1}{\cos^2(x)}$.

$\frac{1}{\cos^2(x)}>1$ is true for the given interval $0<x<\frac{pi}{2}$

We can deduce from the MVT that $\frac{1}{\cos^2(x)}=\frac{\tan(x)-\tan(0)}{x-0}>1$

Therefore:

$$\frac{1}{\cos^2(x)}>1 \implies\text{ with MVT }\frac{\tan(x)-\tan(0)}{x-0}>1$$

That was basically my idea to prove it. What I don't understand, where have I shown, that for every x in the interval $\tan(x)>x$. Suppose we would not know that $\tan(x)>x$ on the given interval is true. Couldn't it be that there are still some x values that are bigger than $\tan(x)$? I mean where have we eliminated this? Why exactly are we using the derivative or why are we using it? Especially if it will be only true for maybe one point in the interval?

And where would the MVT give us an error if we would try to prove an inequality which IS WRONG?

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  • $\begingroup$ The argument is the other way around: One “knows” that $0 < \cos(c) < 1$ for $0 < c < \pi/2$. Therefore $\frac{\tan(x)-\tan(0)}{x-0} = \frac{1}{\cos^2(c)} > 1$. $\endgroup$ – Martin R Jan 18 at 13:09
  • $\begingroup$ @MartinR for clarification reasons. For every c in the interval $0<c<\frac{pi}{2}$? And what I don't get. As far as I know $\frac{1}{cos^2(c)} = \frac{tan(x)-tan(0)}{x-0}$ this equivalence does not have to be satisfied on the whole interval. Therefore how will it helps us if we know there is at least one c which satisfy the equation above but maybe there are some which won't satisfy the equivalence. $\endgroup$ – BrainOverflow Jan 18 at 13:41
  • $\begingroup$ For a given $x \in (0, \pi/2)$ there is (at least) one $c \in (0, x)$ such that $\frac{\tan(x)-\tan(0)}{x-0} = \frac{1}{\cos^2(c)}$. You also know that $ \frac{1}{\cos^2(c)} > 1$. It follows that $\tan(x) > x$. $\endgroup$ – Martin R Jan 18 at 13:47
  • $\begingroup$ @MartinR okay that helped me a lot! Idk why I am struggling with the MVT I get the idea which ain't a problem just in combination with inequality there is smth which boggles my mind. imgur.com/pZ4pl2i In this picture the equation f(x)>g(x) for all x on the red interval would be obv. wrong. Would the MVT fail there If I would try to prove a wrong inequality? $\endgroup$ – BrainOverflow Jan 18 at 14:42
  • $\begingroup$ That question does not make much sense to me. Of course a correct theorem (the MVT), when applied correctly, cannot prove something which is wrong. $\endgroup$ – Martin R Jan 18 at 14:45
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If $f\colon[a,b]\to \Bbb R$ is continuously differentiable, then $\frac{f(b)-f(a)}{b-a}=f'(c)$ for some $c\in [a,b]$. By the extreme value theorem, $f'$ is bounded by some $M$ on $[a,b]$. Since $f(b)-f(a)=(b-a)f'(c)$, we have $|f(b)-f(a)|\le (b-a)M$. That's why the french call it the Théorème des accroissements finis .

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  • $\begingroup$ okay it is bounded but I don't get how it answeres my question. I mean it says there is a point c in the Interval with $f'(c)=\frac{tan(x)-tan(0)}{x-0}$ But how does it helps us with the inequality hence there is probably only one c which satisfies the equation. $\endgroup$ – BrainOverflow Jan 18 at 13:05

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