Question

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

Explanation:

Step1: Find derivative of $f(x)$

Use chain rule: $f'(x)=4(x^2-1)^3\cdot 2x = 8x(x^2-1)^3$

Step2: Find critical points

Set $f'(x)=0$:
$8x(x^2-1)^3=0$
Solutions: $x=-1,0,1$

Step3: Test sign of $f'(x)$

• For $x < -1$: $x<0$, $(x^2-1)^3>0$, so $f'(x)<0$ (function increasing)
• For $-1 < x < 0$: $x<0$, $(x^2-1)^3<0$, so $f'(x)>0$ (function decreasing)
• For $0 < x < 1$: $x>0$, $(x^2-1)^3<0$, so $f'(x)<0$ (function increasing)
• For $x > 1$: $x>0$, $(x^2-1)^3>0$, so $f'(x)>0$ (function decreasing)

Step4: Identify decreasing intervals

Intervals where $f'(x)<0$: $(-1,0)$ and $(1,\infty)$

Answer:

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