independent practice\ndetermine the intervals of decrease of the polynomial function.\nf(x)=…

Explanation:

Step1: Find the derivative

$f'(x)=-12x^{3}-48x^{2}+12x + 48$

Step2: Factor the derivative

$f'(x)=-12(x^{3}+4x^{2}-x - 4)=-12[x^{2}(x + 4)-(x + 4)]=-12(x + 4)(x^{2}-1)=-12(x + 4)(x - 1)(x+1)$

Step3: Determine critical points

Set $f'(x)=0$, then $x=-4,x = 1,x=-1$

Step4: Test intervals

Test the intervals $(-\infty,-4),(-4,-1),(-1,1),(1,\infty)$ using test - points. For $x<-4$, let $x=-5$, $f'(-5)=-12(-5 + 4)(-5 - 1)(-5+1)<0$ For $-4<x<-1$, let $x = - 2$, $f'(-2)=-12(-2 + 4)(-2 - 1)(-2+1)>0$ For $-1<x<1$, let $x = 0$, $f'(0)=-12(0 + 4)(0 - 1)(0+1)>0$ For $x>1$, let $x = 2$, $f'(2)=-12(2 + 4)(2 - 1)(2+1)<0$ The function is decreasing when $f'(x)<0$.

Answer:

$(-\infty,-4)\cup(1,\infty)$