Question

lf $a<0$ the function $e^{ax}+e^{-ax}$ is a monotonically decreasing function for values of $x$ given by

• A. $x>0$
• B. $x<0$
• C. $x>1$
• D. $x<1$

Answer 1

The correct option is B $x<0$

$f\left(x\right)=e^{ax}+e^{-ax}$
$f^{\prime }\left(x\right)=a(e^{ax}-e^{-ax})$
$f^{\prime }\left(x\right)<0$ (for decreasing $f\left(x\right)$)
$a(e^{ax}-e^{-ax})<0$
$e^{ax}-e^{-ax}>0$ (as $a<0)$
$e^{ax}>e^{-ax}$
$e^{2ax}>1$
So $x<0$