JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Application of Derivatives

$f(x)=(x+3{)}^2(x-2{)}^3$.

$f^{\prime }(x)=2(x+3)(x-2{)}^3+3(x+3{)}^2(x-2{)}^2=(x+3)(x-2{)}^2[2(x-2)+3(x+3)]=5(x+3)(x+1)(x-2{)}^2$.

Since $(x-2{)}^2\ge 0$, the sign of $f^{\prime }$ is controlled by $(x+3)(x+1)$: positive on $(-4,-3)$, negative on $(-3,-1)$, positive on $(-1,4)$ (passing through $x=2$, where $f^{\prime }$ touches zero without changing sign).

So $f$ has a local maximum at $x=-3$ and a local minimum at $x=-1$; comparing these with the endpoints gives the global extrema on $[-4,4]$.

$f(-4)=(-1{)}^2(-6{)}^3=-216$, $f(-3)=0$, $f(-1)=(2{)}^2(-3{)}^3=-108$, $f(2)=0$, $f(4)=(7{)}^2(2{)}^3=392$.

So $M=392$, $m=-216$, and $M-m=608$.