Question

Find the maximum value of $2x^3-24x+107$ in the interval [1,3]. Find the maximum value of the same function in [-3,-1].

Answer 1

Let $f\left(x\right)=2x^3-24x+107$
$\Rightarrow f^{\prime }\left(x\right)=6x^2-24=6(x^2-4)=6(x+2)(x-2)$
For maxima or minima put $f^{\prime }\left(x\right)=0\Rightarrow 6(x+2)(x-2)=0\Rightarrow x=2,-2$
We first consider the interval [1,3].
So, we have to evaluate the value of f at the critical point $x=2ϵ[1,3]$ and at the end points of [1,3].
$\begin{array}{cc}Atx=1,& f\left(1\right)=2\times 1^3-24\times 1+107=85\\ Atx=2,& f\left(2\right)=2\times 2^3-24\times 2+107=75\\ Atx=3,& f\left(3\right)=2\times 3^3-24\times 3+107=89\end{array}$
Hence, the absolute maximum value of f(x) in the interval [1,3] is 89 occuring at x=3.
Next, we consider the function in the interval [-3, -1], then evaluate the value of f at the critical point $x=-2ϵ[-3,-1]$ and at the end points of the interval [-3, -1]
$\begin{array}{cc}Atx=1,& f(-1)=2(-1{)}^3-24(-1)+107=-2+24+107=129\\ Atx=2,& f(-2)=2(-2{)}^3-24(-2)+107=-16+48+107=139\\ Atx=3,& f(-3)=2(-3{)}^3-24(-3)+107=-54+72+107=125\end{array}$
Hence, the absolute maximum value of f(x) in the interval [-3, -1] is 139 occuring at x=-2