Answer
(a) $f(x)$ is a one-to-one function.
(b) $f^{-1}(1) = 0$
(c) $(f^{-1})' (1) = \frac{1}{2}$
Work Step by Step
(a) Choose any two real numbers $x_1$ and $x_2$.
Suppose that $x_1 \gt x_2$
Then $~~~e^{x_1} \gt e^{x_2}$
Thus:
$x_1 + e^{x_1} \gt x_2 + e^{x_2}$
Therefore, $f(x)$ is a one-to-one function.
(b) Suppose $f(x) = 1$
$x+e^x = 1$
Then $x = 0$ since $0+e^0 = 1$
Therefore:
$f^{-1}(1) = 0$
(c) $f(x) = x+e^x$
$f'(x) = 1+e^x$
We can find $(f^{-1})' (1)$:
$(f^{-1})' (1) = \frac{1}{f'(f^{-1}(1))}$
$(f^{-1})' (1) = \frac{1}{f'(0)}$
$(f^{-1})' (1) = \frac{1}{1+e^0}$
$(f^{-1})' (1) = \frac{1}{1+1}$
$(f^{-1})' (1) = \frac{1}{2}$
