Question

let f be defined by f(x)=(x² - 1)⁴ for all real numbers x. for what values of x is the function increasing? (-1,1) (-1,0) and (1,∞) (1,∞) (-∞,-1) and (1,∞)

Explanation:

Step1: Differentiate the function

Use the chain - rule. If \(y = u^{n}\) where \(u=x^{2}-1\) and \(n = 4\), then \(y^\prime=nu^{n - 1}\cdot u^\prime\).
First, \(u^\prime=\frac{d}{dx}(x^{2}-1)=2x\).
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 critical points

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

Step3: Use the test - intervals

We have the intervals \((-\infty,-1),(-1,0),(0,1),(1,\infty)\).

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

Answer:

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