let $f(x) = (x^2 - 1)^4$. over what interval is the function decreasing?\\n\\(\\bigcirc\\) $(1…

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

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

Answer

Answer:

$(-1,0)$ 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: $x=0$, $x=1$, $x=-1$

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

  • For $x < -1$: $x<0$, $(x^2-1)^3>0$, so $f'(x)<0$? No, $8x(x^2-1)^3 = 8*(negative)*(positive)=negative$? Correction: Wait, $x^2-1$ when $x<-1$ is positive, cubed is positive, $x$ is negative, so $f'(x)=negative$. But wait, function increases when $f'(x)>0$, decreases when $f'(x)<0$.
  • For $-1<x<0$: $x<0$, $(x^2-1)^3<0$, so $f'(x)=8*(negative)(negative)=positive$? No, wait $(x^2-1)$ is negative when $|x|<1$, cubed is negative. $x$ is negative, so $negative * negative = positive$? No, $8x(x^2-1)^3 = 8(-)(-)=+$, so $f'(x)>0$? No, wait no: $x$ in $(-1,0)$ is negative, $(x^2-1)$ is negative, cubed is negative. So $8(-)(-)=+$, so $f'(x)>0$? Wait no, wait $f(x)=(x^2-1)^4$. Let's test $x=-0.5$: $f(-0.5)=((0.25)-1)^4=(-0.75)^4=0.316$. $x=0$: $f(0)=1$. So from $-1$ to $0$, function goes from 0 to 1, increasing. Then $x$ in $(0,1)$: $x>0$, $(x^2-1)^3<0$, so $f'(x)=8(+)(-)=-$, so $f'(x)<0$, function decreasing. $x>1$: $x>0$, $(x^2-1)^3>0$, so $f'(x)=8(+)(+)=+$, no wait no: $x=2$, $f(2)=(4-1)^4=81$, $x=1$ is 0, so from 1 to infinity, function goes from 0 to 81, increasing? Wait no, wait I messed up the derivative. Wait chain rule: $d/dx (u^4)=4u^3 u'$, $u=x^2-1$, $u'=2x$. So $f'(x)=4(x^2-1)^32x=8x(x^2-1)^3$. Correct.

Wait let's re-test intervals:

  1. $(-\infty,-1)$: Let $x=-2$, $f'(-2)=8*(-2)((4-1)^3)= -1627=-432 <0$, so function decreasing here.
  2. $(-1,0)$: $x=-0.5$, $f'(-0.5)=8*(-0.5)((0.25-1)^3)= -4(-0.75)^3= -4*(-0.421875)=1.6875>0$, function increasing here.
  3. $(0,1)$: $x=0.5$, $f'(0.5)=8*(0.5)((0.25-1)^3)=4(-0.421875)=-1.6875<0$, function decreasing here.
  4. $(1,\infty)$: $x=2$, $f'(2)=82(3)^3=16*27=432>0$, function increasing here.

Wait wait the question is asking where the function is decreasing. So intervals where $f'(x)<0$ are $(-\infty,-1)$ and $(0,1)$? But that's option C. Wait no, wait let's compute $f(-2)=((4)-1)^4=81$, $f(-1)=0$, so from $-\infty$ to $-1$, function goes from 81 to 0, decreasing. $f(0)=1$, $f(1)=0$, so from 0 to1, function goes from 1 to0, decreasing. $f(1)=0$, $f(2)=81$, increasing. $f(-1)=0$, $f(0)=1$, increasing. Oh! I made a mistake earlier in evaluating $f(-0.5)$: $f(-0.5)=((-0.5)^2-1)^4=(0.25-1)^4=(-0.75)^4=0.316$, which is less than $f(0)=1$, so yes, from -1 to 0, function increases from 0 to1. From 0 to1, function decreases from1 to0. From 1 to infinity, increases from0 to infinity. From -infinity to -1, decreases from infinity to0.

Wait but the options: Option C is $(-\infty,-1)$ and $(0,1)$, which are the decreasing intervals. Wait but let me check again.

Wait $f(x)=(x^2-1)^4$ is a even function, symmetric about y-axis. So left side mirrors right side. For $x>0$:

  • $0<x<1$: $x^2-1$ is negative, raised to 4 is positive, as x increases from 0 to1, $x^2-1$ goes from -1 to0, so $(x^2-1)^4$ goes from1 to0, decreasing.
  • $x>1$: $x^2-1$ goes from0 to infinity, so $(x^2-1)^4$ goes from0 to infinity, increasing.

For $x<0$:

  • $x<-1$: $x^2-1$ goes from infinity to0, so $(x^2-1)^4$ goes from infinity to0, decreasing.
  • $-1<x<0$: $x^2-1$ goes from0 to-1, so $(x^2-1)^4$ goes from0 to1, increasing.

So the function is decreasing on $(-\infty,-1)$ and $(0,1)$, which is option C. Wait but I messed up the derivative sign earlier for $x<-1$: $x$ is negative, $(x^2-1)^3$ is positive (since $x^2-1>0$), so $8x(x^2-1)^3$ is negative, so $f'(x)<0$, correct, function decreasing. For $0<x<1$: $x$ positive, $(x^2-1)^3$ negative, so $f'(x)$ negative, function decreasing.

Yes, so the correct interval is $(-\infty,-1)$ and $(0,1)$, which is option C.

Wait wait no, let's check the options again: Options:

  1. $(1,\infty)$
  2. $(-1,0)$
  3. $(-\infty,-1)$ and $(0,1)$
  4. $(-1,0)$ and $(1,\infty)$

Yes, option 3 is $(-\infty,-1)$ and $(0,1)$, which are the decreasing intervals.

Step4: Confirm decreasing intervals

Function decreases when $f'(x)<0$, which occurs on $(-\infty,-1)$ and $(0,1)$