Question

The minimum value of the function f(x) = 2x³$-$21x²+36x$-$20 is

(a) $-$128

(b) $-$126

(c) $-$120

(d) none of these

Answer 1

(a)$-$128

$$\begin{gathered} \mathrm{Given}:\hspace{0.17em}f\left(x\right)=2x^3-21x^2+36x-20 \\ \Rightarrow f\text{'}\left(x\right)=6x^2-42x+36 \\ \mathrm{For}\mathrm{a}\mathrm{local}\mathrm{maxima}\mathrm{or}\mathrm{a}\mathrm{local}\mathrm{minima},\mathrm{we}\mathrm{must}\mathrm{have} \\ f\text{'}\left(x\right)=0 \\ \Rightarrow 6x^2-42x+36=0 \\ \Rightarrow x^2-7x+6=0 \\ \Rightarrow \left(x-1\right)\left(x-6\right)=0 \\ \Rightarrow x=1,6 \\ \mathrm{Now}, \\ f\text{'}\text{'}\left(x\right)=12x-42 \\ \Rightarrow f\text{'}\text{'}\left(1\right)=12-42=-30<0 \\ \mathrm{So},x=1\mathrm{is}\mathrm{a}\mathrm{local}\mathrm{maxima}. \\ \mathrm{Also}, \\ f\text{'}\text{'}\left(6\right)=72-42=30>0 \\ \mathrm{So},x=6\mathrm{is}\mathrm{a}\mathrm{local}\mathrm{miniima}. \\ \mathrm{The}\mathrm{local}\mathrm{minimum}\mathrm{value}\mathrm{is}\mathrm{given}\mathrm{by} \\ f\left(6\right)==2{\left(6\right)}^3-21{\left(6\right)}^2+36\left(6\right)-20=-128 \end{gathered}$$