Question

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? $-\infty2$ 3 multiple choice 4 points let $f(x)=(x^{2}-1)^{3}$. over what interval is the function decreasing? $(-1,0$ $(-\infty,0)$ $(-1,1)$ $(1,\infty)$

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
• When \(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
• 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

eq - 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. \((-\infty,0)\)