Question

For the function $f\left(x\right)=\frac{e^{x^2}-e^{x^{-2}}}{e^{x^2}+e^{x^{-2}}},$ which of the following is true regarding its monotonicity at $x=3$ is

• A. Increasing
• B. Decreasing
• C. Non-monotonic
• D. Cannot be determined

Answer 1

The correct option is A Increasing

$f\left(x\right)=\frac{e^{x^2}-e^{x^{-2}}}{e^{x^2}+e^{x^{-2}}}$
$f\left(x\right)=\frac{e^{x^2}+e^{x^{-2}}-2e^{x^{-2}}}{e^{x^2}+e^{x^{-2}}}$
$\Rightarrow f\left(x\right)=1-\frac{2e^{x^{-2}}}{e^{x^2}+e^{x^{-2}}}$
$\Rightarrow f\left(x\right)=1-\frac{2e^{1/x^2}}{e^{x^2}+e^{1/x^2}}$
$\Rightarrow f\left(x\right)=1-\frac{2}{e^{x^2-\frac{1}{x^2}}+1}$
$\Rightarrow f\left(x\right)=1-\frac{2}{e^{\frac{x^4-1}{x^2}}+1}$

$f^{\prime }\left(x\right)=\frac{2(4x-\frac{2(x^4-1)}{x^3})e^{\frac{x^4-1}{x^2}}}{{(e^{\frac{x^4-1}{x^2}}+1)}^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{4(x^4+1)e^{\frac{x^4-1}{x^2}}}{x^3{(e^{\frac{x^4-1}{x^2}}+1)}^2}$
Clearly $f^{\prime }\left(3\right)>0$
$\Rightarrow f\left(x\right)$ is increasing.