Question

$f\left(x\right)=x+\frac{1}{x},x\ne 0,$, then

• A. $f\left(x\right)$ has no point of local maxima
• B. $f\left(x\right)$ has no point local minima
• C. $f\left(x\right)$ has exactly has one point of local minima
• D. $f\left(x\right)$ has exactly has two point of local minima

Answer 1

The correct option is C $f\left(x\right)$ has exactly has one point of local minima

$f\left(x\right)=x+\frac{1}{x}x\ne 0$

graph of $f\left(x\right)\to$

$f^{\prime }\left(x\right)=1-\frac{1}{x^2}=0$

$x=\pm 1$

$f"\left(x\right)=\frac{2}{x^3}$ can $k+ive$ on $x-ive\pm 1$

There is are point of local minima & point of local maxima

$C$ is correct.