Question

let ( f ) be defined by ( f(x)=(x^{2}-1)^{4} ) for all real numbers ( x ). for what values of ( x ) is the function increasing?
( (-1,1) )
( (-1,0) ) and ( (1, infty) )
( (-infty,-1) ) and ( (1, infty) )
( (1, infty) )

Explanation:

Step1: Differentiate the function

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 \(f^\prime(x)=4(x^{2}-1)^{3}\cdot2x=8x(x^{2}-1)^{3}=8x(x - 1)^{3}(x + 1)^{3}\)

Step2: Find critical points

Set \(f^\prime(x)=0\). Then \(8x(x - 1)^{3}(x + 1)^{3}=0\). The critical points are \(x=-1,x = 0,x = 1\)

Step3: Test intervals

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

A function \(y = f(x)\) is increasing when \(f^\prime(x)>0\). So \(f(x)\) is increasing on the intervals \((-1,0)\) and \((1,\infty)\)

Answer:

B. \((-1,0)\) and \((1,\infty)\)