Question

Construct the graph of the given function and state its nature:

$y=\frac{e^x}{x}$

• A. Increasing for $(-\infty ,0)$ and decreasing for $(0,\infty )$
• B. Increasing for $(1,\infty )$ and decreasing for $(-\infty ,1)-\left\{0\right\}$
  
• C. Increasing for $(-1,\infty )-\left\{0\right\}$ and decreasing for $(-\infty ,1)$
• D. Increasing for $(-\infty ,1)-\left\{0\right\}$ and decreasing for $(1,\infty )$

Answer 1

The correct option is B Increasing for $(1,\infty )$ and decreasing for $(-\infty ,1)-\left\{0\right\}$

Given function is $y=\frac{e^x}{x}$
$y=0$ implies $e^x=0$
Hence, there is no solution.
Hence, $y=f\left(x\right)$ does not cut the $x$-axis at any point.
${lim}_{x\to 0^{+}}$ $=\frac{e^x-1}{x}+\frac{1}{x}$ $=\infty$
Similarly ${lim}_{x\to 0^{-}}$ $=\frac{e^x-1}{x}+\frac{1}{x}$ $=-\infty$
Hence, discontinuous at $x=0$.
Now $f^{\prime }\left(x\right)>0$ implies $x\left(e^x\right)-e^x>0$
$\Rightarrow e^x(x-1)>0$
$\Rightarrow x>1$
Hence, function is increasing for $(1,\infty )$ and decreasing for $(-\infty ,1)-\left\{0\right\}$