Calculus about solving equation with exp.

Question

Prove that the equation $e^x − x − 1 = 0$ has one and only solution. Find that solution.

Hello there! I got this problem on my test last month and I could not solve it at the time. However, I would like to know how the problem could be solved. Thanks for the reply.

Answer 1

Set $f(x)=\mathrm e^x-x-1$. Then $f'(x)=\mathrm e^x-1<0$ if $x<0$, $>0$ if $x>0$, i.e. $f(x)$ decreases to $f(0)=0$ on $\;(-\infty,0]$, then increases from $0$ on $\;[0,+\infty)$, so the sole root of $f(x)=0$ is $x=0$.

Answer 2

By inspection $x=0, y=1 $ is a solution.

The derivative of $ e^x$ at $x=0$ is $1$. The derivative of $ x+1$ at $x=0$ is also $1$.

So the straight line has tangential contact, a double or coincident soluton.

Repeated points $ (0,1),(0,1)$ are the required double contact points.