Answer
$f(x)=e^{x}-x-1$
Work Step by Step
The tangent line at $(x, f(x))$ has slope $f'(x)$, which is given as:
$f'(x)=e^{x}-1$
So, $f(x)$ is an antiderivative:
$f(x)=\displaystyle \int(e^{x}-1)dx=e^{x}-x+C.$
The indefinite integral gives us a collection of functions.
To find the exact function, we find C.
We are given that $f(0)=0$, so
$0=e^{0}-0+C$
$0=1+C$
$C=-1$
So,
$f(x)=e^{x}-x-1.$
