Question

Let $f\left(x\right)=\frac{x-1}{x^2}$ then which of the following is correct.

• A. $f\left(x\right)$ has minima but no maxima.
• B. $f\left(x\right)$ increases in the interval $(0,2)$ and decreases in the interval $(-\infty ,0)\cup (2,\infty )$.
• C. $f\left(x\right)$ is concave down in $(-\infty ,0)\cup (0,3)$.
• D. $x=3$ is the point of inflection.

Answer 1

Answer: (B), (C), (D)

The correct options are
B $f\left(x\right)$ increases in the interval $(0,2)$ and decreases in the interval $(-\infty ,0)\cup (2,\infty )$.
C $x=3$ is the point of inflection.
D $f\left(x\right)$ is concave down in $(-\infty ,0)\cup (0,3)$.

$f\left(x\right)=\frac{x-1}{x^2}$
$f^{\prime }\left(x\right)=\frac{x^2-(x-1).2x}{(x^4}=\frac{2-x}{x^3}$
For $f\left(x\right)$ to be increasing
$f^{\prime }\left(x\right)>0\Rightarrow \frac{2-x}{x^3}>0$
$\Rightarrow xϵ(0,2)$
For $f\left(x\right)$ to be decreasing
$f^{\prime }\left(x\right)<0\Rightarrow \frac{2-x}{(x^3}<0$
$\Rightarrow xϵ(-\infty ,0)\cup (2,\infty )$
Now $f^{\prime \prime }\left(x\right)=\frac{x^3(-1)-(2-x).3x^2}{x^6}=\frac{2(x-3)}{x^4}$
For $f\left(x\right)$ to be concave down $f^{\prime \prime }\left(x\right)<0$
$\Rightarrow \frac{2(x-3)}{x^4}<0\Rightarrow xϵ(-\infty ,0)\cup (3,\infty )$
also for point of inflection $f^{\prime \prime }\left(x\right)=0\Rightarrow x=3$