Question

Find the maximum and minimum values of the function f(x) = $\frac{4}{x+2}+x.$

Answer 1

$$\begin{gathered} \mathrm{Given}:f\left(x\right)=\frac{4}{x+2}+x \\ \Rightarrow f\text{'}\left(x\right)=-\frac{4}{{\left(x+2\right)}^2}+1 \\ \mathrm{For}\mathrm{a}\mathrm{local}\mathrm{maxima}\mathrm{or}\mathrm{a}\mathrm{local}\mathrm{minima},\mathrm{we}\mathrm{must}\mathrm{have} \\ f\text{'}\left(x\right)=0 \\ \Rightarrow -\frac{4}{{\left(x+2\right)}^2}+1=0 \\ \Rightarrow -\frac{4}{{\left(x+2\right)}^2}=-1 \\ \Rightarrow {\left(x+2\right)}^2=4 \\ \Rightarrow x+2=\pm 2 \\ \Rightarrow x=0\mathrm{and}-4 \\ \mathrm{Thus},x=0\mathrm{and}x=-4\mathrm{are}\mathrm{the}\mathrm{possible}\mathrm{points}\mathrm{of}\mathrm{local}\mathrm{maxima}\mathrm{or}\mathrm{local}\mathrm{minima}. \\ \mathrm{Now}, \\ f\text{'}\text{'}\left(x\right)=\frac{8}{{\left(x+2\right)}^3} \\ \mathrm{At}x=0: \\ f\text{'}\text{'}\left(0\right)=\frac{8}{{\left(2\right)}^3}=1>0 \\ \mathrm{So},x=0\mathrm{is}\mathrm{a}\mathrm{point}\mathrm{of}\mathrm{local}\mathrm{minimum}. \\ \mathrm{The}\mathrm{local}\mathrm{minimum}\mathrm{value}\mathrm{is}\mathrm{given}\mathrm{by} \\ f\left(0\right)=\frac{4}{0+2}+0=2 \\ \mathrm{At}x=-4: \\ f\text{'}\text{'}\left(-4\right)=\frac{8}{{\left(-4\right)}^3}=\frac{-1}{8}<0 \\ \mathrm{So},x=-4\mathrm{is}\mathrm{a}\mathrm{point}\mathrm{of}\mathrm{local}\mathrm{minimum}. \\ \mathrm{The}\mathrm{local}\mathrm{maximum}\mathrm{value}\mathrm{is}\mathrm{given}\mathrm{by} \\ f\left(-4\right)=\frac{4}{-4+2}-4=-6 \end{gathered}$$