Answer
$f(x)=2e^{x}+x-2e-2$
Work Step by Step
The tangent line at $(x, f(x))$ has slope $f'(x)$, which is given as:
$f'(x)=2e^{x}+1$
So, $f(x)$ is an antiderivative:
$f(x)=\displaystyle \int(2e^{x}+1)dx=2e^{x}+x+C.$
The indefinite integral gives us a collection of functions.
To find the exact function, we find C.
We are given that $f(1)=-1$, so
$-1=2e^{1}+1+C$
$-2e-2=C$
So,
$f(x)=2e^{x}+x-2e-2$