Question

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

Explanation:

Step1: Find the derivative of \(f(x)\)

Use the chain rule. If \(y = u^n\) where \(u=x^{2}-1\) and \(n = 4\), then \(y^\prime=nu^{n - 1}\cdot u^\prime\).
\(u^\prime=2x\), so \(f^\prime(x)=4(x^{2}-1)^{3}\cdot2x=8x(x^{2}-1)^{3}\).
Factor \(x^{2}-1=(x + 1)(x - 1)\), then \(f^\prime(x)=8x(x + 1)^{3}(x - 1)^{3}\).

Step2: Find the critical points

Set \(f^\prime(x)=0\).
\(8x(x + 1)^{3}(x - 1)^{3}=0\) gives \(x=-1,x = 0,x = 1\).

Step3: Test intervals

• For \(x\in(-\infty,-1)\), let \(x=-2\). Then \(f^\prime(-2)=8\times(-2)\times(-2 + 1)^{3}\times(-2 - 1)^{3}=8\times(-2)\times(-1)^{3}\times(-3)^{3}=8\times(-2)\times(-1)\times(-27)=-432<0\).
• For \(x\in(-1,0)\), let \(x =-\frac{1}{2}\). Then \(f^\prime(-\frac{1}{2})=8\times(-\frac{1}{2})\times(-\frac{1}{2}+1)^{3}\times(-\frac{1}{2}-1)^{3}=8\times(-\frac{1}{2})\times(\frac{1}{2})^{3}\times(-\frac{3}{2})^{3}=8\times(-\frac{1}{2})\times\frac{1}{8}\times(-\frac{27}{8})=\frac{27}{16}>0\).
• For \(x\in(0,1)\), let \(x=\frac{1}{2}\). Then \(f^\prime(\frac{1}{2})=8\times\frac{1}{2}\times(\frac{1}{2}+1)^{3}\times(\frac{1}{2}-1)^{3}=8\times\frac{1}{2}\times(\frac{3}{2})^{3}\times(-\frac{1}{2})^{3}=4\times\frac{27}{8}\times(-\frac{1}{8})=-\frac{27}{16}<0\).
• For \(x\in(1,\infty)\), let \(x = 2\). Then \(f^\prime(2)=8\times2\times(2 + 1)^{3}\times(2 - 1)^{3}=16\times27\times1 = 432>0\).

A function \(y = f(x)\) is decreasing when \(f^\prime(x)<0\).

Answer:

\((-\infty,-1)\) and \((0,1)\) (the third option)