Question

The $f\left(x\right)=(3-x)e^{2x}-4x{e}^x-x$ has

• A. a maximum at $x=0$
• B. a minimum at $x=0$
• C. neither of two at $x=0$
• D. $f\left(x\right)$ is not differentiable at $x=0$

Answer 1

The correct option is C neither of two at $x=0$

$f\left(x\right)=(3-x)e^{2x}-4x{e}^x-x$
$f^{\prime }\left(x\right)=(3-x)2e^{2x}-e^{2x}-4x{e}^x-4e^x-1=(5-2x)3^{2x}-(4x+4)e^x-1$
$f^{\prime \prime }\left(x\right)=(-2)2e^{2x}+(5-2x)2e^{2x}-(4x+4)e^x-4e^x$
Clearly $f^{\prime }\left(0\right)=0=f^{\prime \prime }\left(0\right)$
Hence at $x=0$ f will neither attain it's maxima or minima.