Find $x\in \mathbb{R}$ so $e^{x}-1=x(e-1)$

Question

Find $x\in \mathbb{R}$ so $e^{x}-1=x(e-1)$

I tried using Fermat theorem, but I'm not sure that $f= e^{x}-1-x(e-1)$is increasing

Answer 1

Let $f(x)=e^x-1-x(e-1).$

Thus, $f''(x)=e^x>0,$ which says that $f$ is a convex function.

Hence, our equation has two roots maximum.

But $0$ an $1$ they are roots and we are done!

Answer 2

You could have checked it easily with graphs . The graph of $y=e^x-1$ goes as

While that of $y=x(e-1)$ goes as

Hence the graph of $e^x-1=x(e-1)$

Note: This could also have been done manually.

Hence our solutions are $x=0,1$