Question

Let $f(x)=(1-1/x)2,x>0$. Then

• A. $f$ is increasing in $(0,2)$ and decreasing in $(2,∞)$
• B. $f$ is decreasing in $(0,2)$ and increasing in $(2,∞)$
• C. $f$ is increasing in $(0,1)$ and decreasing in $(1,∞)$
• D. $f$ is decreasing in $(0,1)$ and increasing in $(1,∞)$
• E. $f$ is increasing in $(0,∞)$

Answer 1

Answer: (D)

$f$ is decreasing in $(0,1)$ and increasing in $(1,∞)$

Answer 2

Given that:

$f(x)=(1-\dfrac{1}{x})^{2}$

$f'(x)=2(1-\dfrac{1}{x})(\dfrac{-1}{x^2})$

now put, $f'(x)=0$

so ,$f'(x)=2(1-\dfrac{1}{x})(\dfrac{-1}{x^2})=0$

$(1-\dfrac{1}{x})(\dfrac{-1}{x^2})=0$

but here for any positive value of $x$, $(\dfrac{-1}{x^2})$ can not be zero

hence , $(1-\dfrac{1}{x})=0$

      $⇒x=1$ ⇢one critical point

Now, on analyzing the intervals based on the behavior of the derivative we can state that,

For$0<x<1,$ $f′(x)<0$. Since $f′(x)$ is negative in this interval, the function f(x) is decreasing in $(0,1)$.

For $x>1$, $f′(x)>0$. Since $f′(x)$ is positive in this interval, the function f(x) is increasing in $(1,∞)$.

Hence , $f$ is decreasing in $(0,1)$ and increasing in $(1,∞)$(_Ans)