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

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,∞)

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,∞)

Answer

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))