Answer
Absolute minimum is $e-2\approx 0.71828$ at $x=1$
Absolute maximum is $1$ at $x=0$
Work Step by Step
$f(x)=e^{x}-2x$
$f'(x)=e^{x}-2$
Critical points:
$f'(x)=0\quad $ when $\quad e^{x}=2\Rightarrow x=\ln 2\approx 0.693$
$f(0)=1-0=1$
$f(\ln 2)=2-2\ln 2\approx 0.6137$
$f(1)=e-2\approx 0.71828$
On $(0,\ln 2)$, testpoint $0.5,\qquad f'(0.5)=-0.35...$, f decreases
On $(\ln 2,1), $ testpoint $0.8,\qquad f'(0.5)=+0.2255...$, f increases
So,
Absolute minimum is at $( 1, e-2)$
Absolute maximum is at $(0,1)$
(and a relative maximum at $(\ln 2, 0.6137)$ )
