Question

Statement I The maximum value of the function $2x^3-24x+107$ in the interval $[1,3]$ is $89$ .
Statement II The maximum value of the function $2x^3-24x+107$ in the interval $[-3,-1]$ is $140$ .

• A. Only statement I is true
• B. Only statement II is true
• C. Both the statements are true
• D. Both the statements are false

Answer 1

Answer: (A)

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