Question

(i) f(x) = 4x $-$ $\frac{x^2}{2}$ in [$-$2,4,5]

(ii) f(x) = (x$-$1)² + 3 in [$-$3,1]

(iii) f(x) = 3x4 $-$ 8x³ + 12x² $-$ 48x + 25 in [0,3]

(iv) f(x) = (x$-$2) $\sqrt{x-1}in[1,9]$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right) \\ \mathrm{Given}:f\left(x\right)=4x-\frac{x^2}{2} \\ \Rightarrow f\text{'}\left(x\right)=4-x \\ \mathrm{For}\mathrm{a}\mathrm{local}\mathrm{maximum}\mathrm{or}\mathrm{a}\mathrm{local}\mathrm{minimum},\mathrm{we}\mathrm{must}\mathrm{have} \\ f\text{'}\left(x\right)=0 \\ \Rightarrow 4-x=0 \\ \Rightarrow x=4 \\ \mathrm{Thus},\mathrm{the}\mathrm{critical}\mathrm{points}\mathrm{of}f\mathrm{are}-2,4\mathrm{and}4.5. \\ \mathrm{Now}, \\ f\left(-2\right)=4\left(-2\right)-\frac{{\left(-2\right)}^2}{2}=-8-2=-10 \\ f\left(4\right)=4\left(4\right)-\frac{{\left(4\right)}^2}{2}=16-8=8 \\ f\left(4.5\right)=4\left(4.5\right)-\frac{{\left(4.5\right)}^2}{2}=18-10.125=7.875 \\ \mathrm{Hence},\mathrm{the}\mathrm{absolute}\mathrm{maximum}\mathrm{value}\mathrm{when}x=4\mathrm{is}8\mathrm{and}\mathrm{the}\mathrm{absolute}\mathrm{minimum}\mathrm{value}\mathrm{when}\mathit{}x=-2\mathrm{is}-10. \end{gathered}$$


$$\begin{gathered} \left(\mathrm{ii}\right) \\ \mathrm{Given}:\hspace{0.17em}f\left(x\right)={\left(x-1\right)}^2+3 \\ \Rightarrow f\text{'}\left(x\right)=2\left(x-1\right) \\ \mathrm{For}\mathrm{a}\mathrm{local}\mathrm{maximum}\mathrm{or}\mathrm{a}\mathrm{local}\mathrm{minimum},\mathrm{we}\mathrm{must}\mathrm{have} \\ f\text{'}\left(x\right)=0 \\ \Rightarrow 2\left(x-1\right)=0 \\ \Rightarrow x=1 \\ \mathrm{Thus},\mathrm{the}\mathrm{critical}\mathrm{points}\mathrm{of}f\mathrm{are}-3\mathrm{and}1. \\ \mathrm{Now}, \\ f\left(-3\right)={\left(-3-1\right)}^2+3=16+3=19 \\ f\left(1\right)={\left(1-1\right)}^2+3=3 \\ \mathrm{Hence},\mathrm{the}\mathrm{absolute}\mathrm{maximum}\mathrm{value}\mathrm{when}x=-3\mathrm{is}19\mathrm{and}\mathrm{the}\mathrm{absolute}\mathrm{minimum}\mathrm{value}\mathrm{when}\mathit{}x=1\mathrm{is}3. \end{gathered}$$


$$\begin{gathered} \left(\mathrm{iii}\right) \\ \mathrm{Given}:\hspace{0.17em}f\left(x\right)=3x^4-8x^3+12x^2-48x+25 \\ \Rightarrow f\text{'}\left(x\right)=12x^3-24x^2+24x-48 \\ \mathrm{For}\mathrm{a}\mathrm{local}\mathrm{maximum}\mathrm{or}\mathrm{a}\mathrm{local}\mathrm{minimum},\mathrm{we}\mathrm{must}\mathrm{have} \\ f\text{'}\left(x\right)=0 \\ \Rightarrow 12x^3-24x^2+24x-48=0 \\ \Rightarrow x^3-2x^2+2x-4=0 \\ \Rightarrow x^2\left(x-2\right)+2\left(x-2\right)=0 \\ \Rightarrow \left(x-2\right)\left(x^2+2\right)=0 \\ \Rightarrow x-2=0\mathrm{or}\left(x^2+2\right)=0 \\ \Rightarrow x=2 \\ \mathrm{No}\mathrm{real}\mathrm{root}\mathrm{exists}\mathrm{for}\left(x^2+2\right)=0. \\ \mathrm{Thus},\mathrm{the}\mathrm{critical}\mathrm{points}\mathrm{of}f\mathrm{are}0,2\mathrm{and}3. \\ \mathrm{Now}, \\ f\left(0\right)=3{\left(0\right)}^4-8{\left(0\right)}^3+12{\left(0\right)}^2-48\left(0\right)+25=25 \\ f\left(2\right)=3{\left(2\right)}^4-8{\left(2\right)}^3+12{\left(2\right)}^2-48\left(2\right)+25=-39 \\ f\left(3\right)=3{\left(3\right)}^4-8{\left(3\right)}^3+12{\left(3\right)}^2-48\left(3\right)+25=16 \\ \mathrm{Hence},\mathrm{the}\mathrm{absolute}\mathrm{maximum}\mathrm{value}\mathrm{when}x=0\mathrm{is}25\mathrm{and}\mathrm{the}\mathrm{absolute}\mathrm{minimum}\mathrm{value}\mathrm{when}x=2\mathrm{is}-39. \end{gathered}$$


$$\begin{gathered} \left(\mathrm{iv}\right) \\ \mathrm{Given}:f\left(x\right)=\left(x-2\right)\sqrt{x-1} \\ \Rightarrow f\text{'}\left(x\right)=\sqrt{x-1}+\frac{\left(x-2\right)}{2\sqrt{x-1}} \\ \mathrm{For}\mathrm{a}\mathrm{local}\mathrm{maximum}\mathrm{or}\mathrm{a}\mathrm{local}\mathrm{minimum},\mathrm{we}\mathrm{must}\mathrm{have} \\ f\text{'}\left(x\right)=0 \\ \Rightarrow \sqrt{x-1}+\frac{\left(x-2\right)}{2\sqrt{x-1}}=0 \\ \Rightarrow 2\left(x-1\right)+\left(x-2\right)=0 \\ \Rightarrow 2x-2+x-2=0 \\ \Rightarrow 3x-4=0 \\ \Rightarrow 3x=4 \\ \Rightarrow x=\frac{4}{3} \\ \mathrm{Thus},\mathrm{the}\mathrm{critical}\mathrm{points}\mathrm{of}f\mathrm{are}1,\frac{4}{3}\mathrm{and}9. \\ \mathrm{Now}, \\ f\left(1\right)=\left(1-2\right)\sqrt{1-1}=0 \\ f\left(\frac{4}{3}\right)=\left(\frac{4}{3}-2\right)\sqrt{\frac{4}{3}-1}=\frac{-2}{3}\times \frac{1}{\sqrt{3}}=-\frac{2}{3\sqrt{3}} \\ f\left(9\right)=\left(9-2\right)\sqrt{9-1}=14\sqrt{2} \\ \mathrm{Hence},\mathrm{the}\mathrm{absolute}\mathrm{maximum}\mathrm{value}\mathrm{when}x=9\mathrm{is}14\sqrt{2}\mathrm{and}\mathrm{the}\mathrm{absolute}\mathrm{minimum}\mathrm{value}\mathrm{when}x=\frac{4}{3}\mathrm{is}-\frac{2}{3\sqrt{3}}. \\ \mathrm{Disclaimer}:\mathrm{The}\mathrm{solution}\mathrm{given}\mathrm{in}\mathrm{the}\mathrm{book}\mathrm{is}\mathrm{incorrect}.\mathrm{The}\mathrm{solution}\mathrm{here}\mathrm{is}\mathrm{created}\mathrm{according}\mathrm{to}\mathrm{the}\mathrm{question}\mathrm{given}\mathrm{in}\mathrm{the}\mathrm{book}. \end{gathered}$$