Question

For the function f(x) = $x+\frac{1}{x}$

(a) x = 1 is a point of maximum
(b) x = $-$1 is a point of minimum
(c) maximum value > minimum value
(d) maximum value< minimum value

Answer 1

$$\begin{gathered} \left(\mathrm{d}\right)\text{maximum value < minimum value} \\ \mathrm{Given}:f\left(x\right)=x+\frac{1}{x} \\ \Rightarrow f\text{'}\left(x\right)=1-\frac{1}{x^2} \\ \mathrm{For}\mathrm{a}\mathrm{local}\mathrm{maxima}\mathrm{or}\mathrm{a}\mathrm{local}\mathrm{minima},\mathrm{we}\mathrm{must}\mathrm{have} \\ f\text{'}\left(x\right)=0 \\ \Rightarrow 1-\frac{1}{x^2}=0 \\ \Rightarrow x^2-1=0 \\ \Rightarrow x^2=1 \\ \Rightarrow x=\pm 1 \\ \mathrm{Now}, \\ f\text{'}\text{'}\left(x\right)=\frac{2}{x^3} \\ \Rightarrow f\text{'}\text{'}\left(1\right)=\frac{2}{1}=2>0 \\ \mathrm{So},x=1\mathrm{is}\mathrm{a}\mathrm{local}\text{minima.} \\ \mathrm{Also}, \\ f\text{'}\text{'}\left(-1\right)=-2<0 \\ \mathrm{So},x=-1\mathrm{is}\mathrm{a}\mathrm{local}\text{maxima.} \\ \mathrm{The}\mathrm{local}\mathrm{minimum}\mathrm{value}\mathrm{is}\mathrm{given}\mathrm{by} \\ f\left(1\right)=2 \\ \mathrm{The}\mathrm{local}\text{maximum}\mathrm{value}\mathrm{is}\mathrm{given}\mathrm{by} \\ f\left(-1\right)=-2 \\ \therefore \mathrm{Maximum\; value}<\mathrm{Minimum}\mathrm{value} \end{gathered}$$