Question

Write the minimum value of f(x) = $x+\frac{1}{x},x>0.$

Answer 1

$$\begin{gathered} \mathrm{Given}:\hspace{0.17em}f\left(x\right)=x+\frac{1}{x} \\ \Rightarrow f\text{'}\left(x\right)=1-\frac{1}{x^2} \\ \mathrm{For}\mathrm{a}\mathrm{local}\mathrm{maxima}\mathrm{or}\mathrm{a}\mathrm{local}\mathrm{minima},\mathrm{we}\mathrm{must}\mathrm{have} \\ f\text{'}\left(x\right)=0 \\ \Rightarrow 1-\frac{1}{x^2}=0 \\ \Rightarrow x^2=1 \\ \Rightarrow x=1,-1 \\ \mathrm{But}x>0 \\ \Rightarrow x=1 \\ \mathrm{Now}, \\ f\text{'}\text{'}\left(x\right)=\frac{1}{x^3} \\ \mathrm{At}x=1: \\ f\text{'}\text{'}\left(1\right)=\frac{2}{{\left(1\right)}^3}=2>0 \\ \mathrm{So},x=1\mathrm{is}\mathrm{a}\mathrm{point}\mathrm{of}\mathrm{local}\mathrm{minimum}. \\ \mathrm{Thus},\mathrm{the}\mathrm{local}\mathrm{minimum}\mathrm{value}\mathrm{is}\mathrm{given}\mathrm{by} \\ f\left(1\right)=1+\frac{1}{1}=1+1=2 \end{gathered}$$