If $f(x)=x+\frac{1}{x}$. Which of the following is not a correct option ?

local maximum value =-2, local minimum value =0

$f(x)=x+\frac{1}{x},\;x\ne0.$

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

$f'(x)=0 \Rightarrow x=\pm1.$

$f(1)=2,\;f(-1)=-2.$

$\text{Local minimum value}=2,\;\text{Local maximum value}=-2.$

$\text{Hence extremes occur at }x=1,-1.$

$\text{Local minimum value is not }0.$

$\text{Incorrect option: local maximum value = −2, local minimum value = 0.}$