## Calculus 8th Edition

$-1$
Need to find the absolute maximum value of the function $f(x) =x-e^{x}$ $f'(x) =1-e^{x}=0$ This implies $e^{x} -1=0$ Thus $x=0$ $f'(x)$ will be increasing for all $x<0$ and $f'(x)$ will be decreasing when $x>0$. Therefore, by first derivative test for absolute extreme values, $f(x) =x-e^{x}$ has an absolute maximum value at x = 0 and absolute maximum value of $f(x) =x-e^{x}$ is $f(0) = 0-1 = -1$ at $x=0$.