Question

Let f(x) = 2x³ $-$ 3x² $-$ 12x + 5 on [$-$2, 4]. The relative maximum occurs at x=

(a) $-$2

(b) $-$1

(c) 2

(d) 4

Answer 1

(c) 2

$$\begin{gathered} \mathrm{Given}:f\left(x\right)=2x^3-3x^2-12x+5 \\ \Rightarrow f\text{'}\left(x\right)=6x^2-6x-12 \\ \mathrm{For}\mathrm{a}\mathrm{local}\mathrm{maxima}\mathrm{or}\mathrm{a}\mathrm{local}\mathrm{minima},\mathrm{we}\mathrm{must}\mathrm{have} \\ f\text{'}\left(x\right)=0 \\ \Rightarrow 6x^2-6x-12=0 \\ \Rightarrow x^2-x-2=0 \\ \Rightarrow \left(x-2\right)\left(x+1\right)=0 \\ \Rightarrow x=2,-1 \\ \mathrm{Now}, \\ f\text{'}\text{'}\left(x\right)=12x-6 \\ \Rightarrow f\text{'}\text{'}\left(-1\right)=-12-6=-18<0 \\ \mathrm{So},x=1\mathrm{is}\mathrm{a}\mathrm{local}\mathrm{maxima}. \\ \mathrm{Also}, \\ f\text{'}\text{'}\left(2\right)=24-6=18>0 \\ \mathrm{So},x=2\mathrm{is}\mathrm{a}\mathrm{local}\mathrm{mini}\mathrm{ma}. \end{gathered}$$