Question

Consider a function $f(x)=x{e}^x+{\left(x{e}^x\right)}^{-1}$ where $x\in R-\{0\}$, then which of the following statement ( s ) is/are INCORRECT?

• A. $f(x)$ attains its local minima at $x=-1$.
• B. $f(x)$ attains its local maxima at $x=\beta$, where $\beta \in (0,1)$.
• C. $f(x)$ attains its local minima at $x=\beta$, where $\beta \in (0,1)$.
• D. $f(x)$ attains its local minima at $x=\beta$, where $\beta \in (1,\infty )$.

Answer 1

Answer: (A), (B), (D)

$f(x)=x{e}^x+\frac{1}{x{e}^x}$ where $x\ne 0$.
$f^{\prime }(x)=\left(x{e}^x+e^x\right)-\frac{1}{{\left(x{e}^x\right)}^2}\left(x{e}^x+e^x\right)=\frac{\left(x{e}^x+e^x\right)\left(x{e}^x-1\right)\left(x{e}^x+1\right)}{{\left(x{e}^x\right)}^2}$
$=\frac{e^{3x}(x+1)\left(x+e^{-x}\right)\left(x-e^{-x}\right)}{{\left(x{e}^x\right)}^2}$
$f^{\prime }(x)=0;x=-1,e^{-x}=-x\text{ (No solution) }$
$e^{-x}=x(\text{ exactly one solution, let it is }\beta )$
$\text{ at }x=-1,\text{ local maxima }$
$\text{ at }x=\beta \in (0,1)\text{ local minima. }$