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In my exam today I had the question with the following inequality, that they wanted us to solve with the mean value theorem.

$$1+x \leq e^x$$

Now, I know the mean value theorem is about

$$ f: [a,b] \rightarrow \mathbb{R}, \ continuous \ on \ [a,b] \ and \ differentiable \ on \ (a,b) \rightarrow \exists \ c \in \mathbb{R} \ such \ that \ f'(c) = \frac{f(b)-f(a)}{b-a}$$

But I had absolutely no clue on how to use this theorem. What I did was first subtracting $e^x$ so the inequality is $1+x-e^x \leq 0$. Then my $f(x) = 1+x-e^x$. They gave us a hint that for $ x < 0: \ \ a:= x \ and \ b:=0 $ would be sensible choices, which I also didn't get. If you differentiate my $f(x)$ and set it equal to $\frac{f(b)-f(a)}{b-a}$, you end up with something like $$x*e^x = 1+x-e^x$$ which I can't solve for the sake of my life.. I know that these two functions only intersect for $x=0$ (at least as long as $ x \in \mathbb{R}$, I think there is some Lambert-W function or something that also solves it?), but I have no clue on how to prove it with the mean-value (Lagrange) theorem.

giuli0110
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  • 1/ Can you show that the derivative is always $\leq 0$? If so, then $f(x) = f'(c) \times (x-0) \leq 0$ by MVT. $\quad$ 2/ To clarify, what is $f'(x)$? You seem to be implying it is just $e^x$? $\quad$ 3/ Avoid having your notation do double meaning. EG In $x \times e^x$, I'm guessing that the $x$'s stand for different things (but also that equation doesn't quite make sense to me per point 2). – Calvin Lin Feb 22 '23 at 13:56
  • There are 33 proofs of that inequality in the duplicate target, including some using the mean-value theorem. – Martin R Feb 22 '23 at 20:53

1 Answers1

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Maybe write what you need in the form of MVT.

To show:

$$\frac{e^x-1}{x}\ge 1$$

Is it more apparent now?

Umesh Shankar
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