Question

The least value of the function f(x) = $x3-18x2+96x$ in the interval [0,9] is
(a) 126
(b) 135
(c) 160
(d) 0

Answer 1

$$\begin{gathered} \left(\mathrm{d}\right)0 \\ \mathrm{Given}:f\left(x\right)=x^3-18x^2+96x \\ \Rightarrow f\text{'}\left(x\right)=3x^2-36x+96 \\ \mathrm{For}\mathrm{a}\mathrm{local}\mathrm{maxima}\mathrm{or}\mathrm{a}\mathrm{local}\mathrm{minima},\mathrm{we}\mathrm{must}\mathrm{have} \\ f\text{'}\left(x\right)=0 \\ \Rightarrow 3x^2-36x+96=0 \\ \Rightarrow x^2-12x+32=0 \\ \Rightarrow \left(x-4\right)\left(x-8\right)=0 \\ \Rightarrow x=4,8 \\ \mathrm{So}, \\ f\left(8\right)={\left(8\right)}^3-18{\left(8\right)}^2+96\left(8\right)=512-1152+768=128 \\ f\left(4\right)={\left(4\right)}^3-18{\left(4\right)}^2+96\left(4\right)=64-288+384=160 \\ f\left(0\right)={\left(0\right)}^3-18{\left(0\right)}^2+96\left(0\right)=0 \\ f\left(9\right)={\left(9\right)}^3-18{\left(9\right)}^2+96\left(9\right)=729-1458+864=135 \\ \text{Hence, 0 is the minimum value in the range}\left[0,9\right]. \end{gathered}$$