1 multiple choice 4 points let $f(x)=(x^{2}-1)^{4}$. over what interval is the function decreasing? $(-1,0)$…

1 multiple choice 4 points let $f(x)=(x^{2}-1)^{4}$. over what interval is the function decreasing? $(-1,0)$ $(1,\\infty)$ $(-\\infty,-1)$ and $(0,1)$ $(-1,1)$ 2 multiple choice 4 points let $f(x)=x^{2}(x - 3)$. over what interval is the function decreasing? $-\\infty<x<\\infty$ $0<x<\\infty$ $0<x<2$ $-\\infty<x<0$ and $x>2$ 3 multiple choice 4 points let $f(x)=(x^{2}-1)^{3}$. over what interval is the function decreasing? $(-1,0$ $(-\\infty,0)$ $(-1,1)$ $(1,\\infty)$

1 multiple choice 4 points let $f(x)=(x^{2}-1)^{4}$. over what interval is the function decreasing? $(-1,0)$ $(1,\\infty)$ $(-\\infty,-1)$ and $(0,1)$ $(-1,1)$ 2 multiple choice 4 points let $f(x)=x^{2}(x - 3)$. over what interval is the function decreasing? $-\\infty<x<\\infty$ $0<x<\\infty$ $0<x<2$ $-\\infty<x<0$ and $x>2$ 3 multiple choice 4 points let $f(x)=(x^{2}-1)^{3}$. over what interval is the function decreasing? $(-1,0$ $(-\\infty,0)$ $(-1,1)$ $(1,\\infty)$

Answer

Explanation:

Step1: Find the derivative of the function

For (y = f(x)=(x^{2}-1)^{4}), using the chain rule ((u^{n})^\prime=nu^{n - 1}\cdot u^\prime), let (u=x^{2}-1), (n = 4). Then (y^\prime=4(x^{2}-1)^{3}\cdot2x=8x(x^{2}-1)^{3}=8x(x - 1)^{3}(x + 1)^{3}). To find where the function is decreasing, we need to find where (y^\prime<0). Set (y^\prime = 0), then (x=-1,0,1). We can use a sign - chart:

  • When (x<-1), let (x=-2), then (y^\prime=8\times(-2)\times((-2)^{2}-1)^{3}=8\times(-2)\times(3)^{3}<0).
  • When (-1<x<0), let (x =-\frac{1}{2}), then (y^\prime=8\times(-\frac{1}{2})\times((-\frac{1}{2})^{2}-1)^{3}=8\times(-\frac{1}{2})\times(-\frac{3}{4})^{3}>0).
  • When (0<x<1), let (x=\frac{1}{2}), then (y^\prime=8\times\frac{1}{2}\times((\frac{1}{2})^{2}-1)^{3}=8\times\frac{1}{2}\times(-\frac{3}{4})^{3}<0).
  • When (x > 1), let (x = 2), then (y^\prime=8\times2\times(2^{2}-1)^{3}=8\times2\times(3)^{3}>0).

For (y = f(x)=x^{2}(x - 3)=x^{3}-3x^{2}), using the power rule ((x^{n})^\prime=nx^{n-1}), (y^\prime=3x^{2}-6x=3x(x - 2)). Set (y^\prime=0), then (x = 0) and (x = 2). Using a sign - chart:

  • When (x<0), let (x=-1), then (y^\prime=3\times(-1)\times(-1 - 2)=9>0).
  • When (0<x<2), let (x = 1), then (y^\prime=3\times1\times(1 - 2)=-3<0).
  • When (x>2), let (x = 3), then (y^\prime=3\times3\times(3 - 2)=9>0).

For (y = f(x)=(x^{2}-1)^{3}), using the chain rule ((u^{n})^\prime=nu^{n - 1}\cdot u^\prime), let (u=x^{2}-1), (n = 3). Then (y^\prime=3(x^{2}-1)^{2}\cdot2x=6x(x^{2}-1)^{2}=6x(x - 1)^{2}(x + 1)^{2}). Set (y^\prime=0), then (x=-1,0,1). Using a sign - chart:

  • When (x<0) ((x\neq - 1)), let (x=-2), then (y^\prime=6\times(-2)\times((-2)^{2}-1)^{2}<0). But when (x=-1), (y^\prime = 0).
  • When (x>0), let (x = 1), (y^\prime=0), let (x = 2), (y^\prime=6\times2\times(2^{2}-1)^{2}>0).

Answer:

  1. ((-\infty,-1)) and ((0,1))
  2. (0 < x < 2)
  3. ((-\infty,0))