Applications of Derivatives
Absolute Extrema on Closed Interval
nta_pyq_2024_jan
Grade 12
Question:
Let $f(x)=(x+3)^2(x-2)^3$, $x\in[-4,4]$. If $M$ and $m$ are the maximum and minimum values of $f$ respectively in $[-4,4]$, then the value of $M-m$ is:
Step-by-Step Solution
Key Concept: $f'(x)=5(x+3)(x-2)^2(x+1)$. Critical points in $[-4,4]$: $x=-3,-1,2$. Evaluate $f$ at critical points and endpoints $x=-4,4$.
$f'(x)=5(x+3)(x-2)^2(x+1)=0$ at $x=-3,-1,2$.
$f(-4)=1\cdot(-216)=-216$, $f(-3)=0$, $f(-1)=4\cdot(-27)=-108$, $f(2)=0$, $f(4)=49\cdot8=392$.
$M=392$, $m=-216$. $M-m=608$.
Correct Answer: 3