Question

consider the function (f(x)=2x^{3}+12x^{2}-72x + 4), (-6leq xleq3). this function has an absolute minimum value equal to and an absolute maximum value equal to question help: video submit question

Explanation:

Step1: Find the derivative

$f'(x)=6x^{2}+24x - 72=6(x^{2}+4x - 12)=6(x + 6)(x - 2)$

Step2: Find critical points

Set $f'(x)=0$, then $6(x + 6)(x - 2)=0$. So the critical points are $x=-6$ and $x = 2$. Both are in the interval $[-6,3]$.

Step3: Evaluate the function at critical and end - points

$f(-6)=2(-6)^{3}+12(-6)^{2}-72(-6)+4=2(-216)+12\times36 + 432+4=-432+432+432+4 = 436$
$f(2)=2(2)^{3}+12(2)^{2}-72(2)+4=2\times8+12\times4-144 + 4=16+48-144 + 4=-76$
$f(3)=2(3)^{3}+12(3)^{2}-72(3)+4=2\times27+12\times9-216+4=54+108-216+4=-50$

Answer:

Absolute minimum value: $-76$
Absolute maximum value: $436$