Showing one function is greater than another function (or proving an inequality)

Question

How would you show that $e^x \lt \frac{1}{1-x} $ for $ 0 \lt x \lt 1 $?

I thought we could define a new function, $h=f-g$, where $f(x)=e^x$ and $g(x)=\frac{1}{1-x}$. Then, move on to calculate the derivative of h and then show that it is negative, implying that h is decreasing. But I did not know how to show that h is negative in $(0,1)$

Does this answer your question? Prove Exponential Function Inequality: $e^x \le \frac{1}{1-x}$ — Martin R, Jun 11 '20 at 07:06

score 2 · Accepted Answer · answered Jun 11 '20 at 01:21

The desired inequality is, of course, equivalent to $$f(x):=e^x(1-x)<1$$ for $0<x<1$. We consider that $$f'(x)=-xe^x\leq0$$ on $[0,1]$, with $f'(x)=0$ exactly when $x=0$. It follows, therefore, that $f$ is monotonically decreasing on $[0,1]$, with a local maximum at $x=0$ where $f(0)=1$.

Showing one function is greater than another function (or proving an inequality)

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