Question

The function $f\left(x\right)=x+\frac{1}{x}$ has

• A. A local maxima at $x=1$ and a local minima at $x=-1$
• B. A local minima at $x=1$ and a local maxima at $x=-1$
• C. Absolute maxima at $x=1$ and absolute minima at $x=-1$
• D. Absolute minima at $x=1$ and absolute maxima at $x=-1$

Answer 1

The correct option is B

A local minima at $x=1$ and a local maxima at $x=-1$

Find the local minima and local maxima:

Given function, $f\left(x\right)=x+\frac{1}{x}$

For critical point $f\text{'}\left(x\right)=0$

For local maxima $f\text{'}\text{'}\left(x\right)<0$ and $f\text{'}\text{'}\left(x\right)>0$

Now

$$\begin{gathered} f\text{'}\left(x\right)=1-\frac{1}{x^2} \\ \Rightarrow 1-\frac{1}{x^2}=0 \\ \Rightarrow x^2-1=0 \\ \Rightarrow x^2=1 \\ \Rightarrow x=\pm 1 \\ f\text{'}\text{'}\left(x\right)=\frac{2}{x^3} \\ f\text{'}\text{'}\left(1\right)=2 \\ f\text{'}\text{'}\left(-1\right)=-2 \end{gathered}$$

Therefore local minima at $x=1$ and a local maxima at $x=-1$

Hence, the correct option is $\left(B\right)$.