Given the function f left parenthesis x right parenthesis equals x superscript 4 baseline plus 6 x cubed minus x squared minus 30 x plus 4, use the Intermediate Value Theorem to decide which of the following intervals contains at least one zero. Select all that apply.

Question

(2 points)
Responses

left bracket negative 5 comma negative 4 right bracket
Image with alt text: left bracket negative 5 comma negative 4 right bracket

left bracket negative 4 comma negative 3 right bracket
Image with alt text: left bracket negative 4 comma negative 3 right bracket

left bracket negative 3 comma negative 2 right bracket
Image with alt text: left bracket negative 3 comma negative 2 right bracket

left bracket negative 1 comma 0 right bracket
Image with alt text: left bracket negative 1 comma 0 right bracket

left bracket 0 comma 1 right bracket
Image with alt text: left bracket 0 comma 1 right bracket

left bracket 1 comma 2 right bracket

Answer 1

To apply the Intermediate Value Theorem (IVT), we need to evaluate the function $f(x)=x^4+6x^3-x^2-30x+4$ at specific points within the provided intervals. The IVT states that if $f(a)$ and $f(b)$ have opposite signs, then there is at least one zero of the function in the interval $(a,b)$.

Let's evaluate the function at the endpoints of each interval:

1. Interval $[-5,-4]$:

   • $f(-5)=(-5{)}^4+6(-5{)}^3-(-5{)}^2-30(-5)+4=625-750-25+150+4=4$
   • $f(-4)=(-4{)}^4+6(-4{)}^3-(-4{)}^2-30(-4)+4=256-384-16+120+4=-20$
   • Since $f(-5)=4>0$ and $f(-4)=-20<0$, the IVT applies. There is at least one zero in $[-5,-4]$.

2. Interval $[-4,-3]$:

   • $f(-4)=-20$ (as calculated above).
   • $f(-3)=(-3{)}^4+6(-3{)}^3-(-3{)}^2-30(-3)+4=81-162-9+90+4=4$
   • Since $f(-4)=-20<0$ and $f(-3)=4>0$, the IVT applies. There is at least one zero in $[-4,-3]$.

3. Interval $[-3,-2]$:

   • $f(-3)=4$ (as calculated above).
   • $f(-2)=(-2{)}^4+6(-2{)}^3-(-2{)}^2-30(-2)+4=16-48-4+60+4=28$
   • Since both $f(-3)=4>0$ and $f(-2)=28>0$, the IVT does not apply here. There is no zero in $[-3,-2]$.

4. Interval $[-1,0]$:

   • $f(-1)=(-1{)}^4+6(-1{)}^3-(-1{)}^2-30(-1)+4=1-6-1+30+4=28$
   • $f(0)=0^4+6(0{)}^3-0^2-30(0)+4=4$
   • Since both $f(-1)=28>0$ and $f(0)=4>0$, the IVT does not apply here. There is no zero in $[-1,0]$.

5. Interval $[0,1]$:

   • $f(0)=4$ (as calculated above).
   • $f(1)=(1{)}^4+6(1{)}^3-(1{)}^2-30(1)+4=1+6-1-30+4=-20$
   • Since $f(0)=4>0$ and $f(1)=-20<0$, the IVT applies. There is at least one zero in $[0,1]$.

6. Interval $[1,2]$:

   • $f(1)=-20$ (as calculated above).
   • $f(2)=(2{)}^4+6(2{)}^3-(2{)}^2-30(2)+4=16+48-4-60+4=4$
   • Since $f(1)=-20<0$ and $f(2)=4>0$, the IVT applies. There is at least one zero in $[1,2]$.

Thus, the intervals that contain at least one zero are:

• $[-5,-4]$
• $[-4,-3]$
• $[0,1]$
• $[1,2]$

In conclusion, the intervals containing at least one zero are:

• $[-5,-4]$
• $[-4,-3]$
• $[0,1]$
• $[1,2]$