let ( f ) be defined by ( f(x)=(x^{2}-1)^{4} ) for all real numbers ( x ). for what values of ( x ) is the…

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

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

Answer

Answer:

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