Let [f(x)=(x+3)^2(x-2)^3, x [-4,4]].

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

Accepted Answer

$$\begin{aligned}
& \mathrm{f}^{\prime}(\mathrm{x})=(\mathrm{x}+3)^2 \cdot 3(\mathrm{x}-2)^2+(\mathrm{x}-2)^3 2(\mathrm{x}+3) \
& =5(\mathrm{x}+3)(\mathrm{x}-2)^2(\mathrm{x}+1) \
& \mathrm{f}^{\prime}(\mathrm{x})=0, \mathrm{x}=-3,-1,2
\end{aligned}$$

$$\begin{aligned}
& f(-4)=-216 \
& f(-3)=0, f(4)=49 \times 8=392 \
& M=392, m=-216 \
& M-m=392+216=608 \
& \text { Ans }=\text { '3' }
\end{aligned}$$

Solutions · 1