Question

Find the absolute maximum and minimum values of a function f given by

$f\left(x\right)=2x^3-15x^2+36x+1\mathrm{on}\mathrm{the}\mathrm{interval}[1,5]$

Answer 1

$$\begin{gathered} \mathrm{Given}:f\left(x\right)=2x^3-15x^2+36x+1 \\ \Rightarrow f\text{'}\left(x\right)=6x^2-30x+36 \\ \mathrm{For}\mathrm{a}\mathrm{local}\mathrm{maximum}\mathrm{or}\mathrm{a}\mathrm{local}\mathrm{minimum},\mathrm{we}\mathrm{have} \\ f\text{'}\left(x\right)=0 \\ \Rightarrow 6x^2-30x+36=0 \\ \Rightarrow x^2-5x+6=0 \\ \Rightarrow \left(x-3\right)\left(x-2\right)=0 \\ \Rightarrow x=2\mathrm{and}x=3 \\ \mathrm{Thus},\mathrm{the}\mathrm{critical}\mathrm{points}\mathrm{of}f\mathrm{are}1,2,3\mathrm{and}5. \\ \mathrm{Now}, \\ f\left(1\right)=2{\left(1\right)}^3-15{\left(1\right)}^2+36\left(1\right)+1=24 \\ f\left(2\right)=2{\left(2\right)}^3-15{\left(2\right)}^2+36\left(2\right)+1=29 \\ f\left(3\right)=2{\left(3\right)}^3-15{\left(3\right)}^2+36\left(3\right)+1=28 \\ f\left(5\right)=2{\left(5\right)}^3-15{\left(5\right)}^2+36\left(5\right)+1=56 \\ \mathrm{Hence},\mathrm{the}\mathrm{absolute}\mathrm{maximum}\mathrm{value}\mathrm{when}x=5\mathrm{is}56\mathrm{and}\mathrm{the}\mathrm{absolute}\mathrm{minimum}\mathrm{value}\mathrm{when}x=1\mathrm{is}24. \end{gathered}$$