Question

given the function $f(x) = x^4 + 6x^3 - x^2 - 30x + 4$, use the intermediate value theorem to decide which of the following intervals contains at least one zero. select all that apply. (2 points) $-5,-4$ $-4,-3$ $-3,-2$ $-1,0$ $0,1$ $1,2$

Explanation:

The Intermediate Value Theorem states that if a function \( f(x) \) is continuous on a closed interval \([a, b]\), and \( k \) is a number between \( f(a) \) and \( f(b) \), then there exists at least one number \( c \) in the interval \((a, b)\) such that \( f(c) = k \). In particular, if \( f(a) \) and \( f(b) \) have opposite signs, then there is at least one zero of the function in the interval \((a, b)\).

We are given the function \( f(x) = x^4 + 6x^3 - x^2 - 30x + 4 \). We need to evaluate \( f(x) \) at the endpoints of each interval and check if the function values have opposite signs.

Step 1: Evaluate \( f(x) \) at \( x = -5 \) and \( x = -4 \)

For \( x = -5 \):
\[

$$\begin{align*} f(-5) &= (-5)^4 + 6(-5)^3 - (-5)^2 - 30(-5) + 4 \\ &= 625 + 6(-125) - 25 + 150 + 4 \\ &= 625 - 750 - 25 + 150 + 4 \\ &= (625 - 750) + (-25 + 150) + 4 \\ &= -125 + 125 + 4 \\ &= 4 \end{align*}$$

\]
For \( x = -4 \):
\[

$$\begin{align*} f(-4) &= (-4)^4 + 6(-4)^3 - (-4)^2 - 30(-4) + 4 \\ &= 256 + 6(-64) - 16 + 120 + 4 \\ &= 256 - 384 - 16 + 120 + 4 \\ &= (256 - 384) + (-16 + 120) + 4 \\ &= -128 + 104 + 4 \\ &= -20 \end{align*}$$

\]
Since \( f(-5) = 4 \) and \( f(-4) = -20 \), and \( 4 \) and \( -20 \) have opposite signs, by the Intermediate Value Theorem, there is at least one zero in the interval \([-5, -4]\).

Step 2: Evaluate \( f(x) \) at \( x = -4 \) and \( x = -3 \)

We already know \( f(-4) = -20 \).
For \( x = -3 \):
\[

$$\begin{align*} f(-3) &= (-3)^4 + 6(-3)^3 - (-3)^2 - 30(-3) + 4 \\ &= 81 + 6(-27) - 9 + 90 + 4 \\ &= 81 - 162 - 9 + 90 + 4 \\ &= (81 - 162) + (-9 + 90) + 4 \\ &= -81 + 81 + 4 \\ &= 4 \end{align*}$$

\]
Since \( f(-4) = -20 \) and \( f(-3) = 4 \), and \( -20 \) and \( 4 \) have opposite signs, there is at least one zero in the interval \([-4, -3]\).

Step 3: Evaluate \( f(x) \) at \( x = -3 \) and \( x = -2 \)

We know \( f(-3) = 4 \).
For \( x = -2 \):
\[

$$\begin{align*} f(-2) &= (-2)^4 + 6(-2)^3 - (-2)^2 - 30(-2) + 4 \\ &= 16 + 6(-8) - 4 + 60 + 4 \\ &= 16 - 48 - 4 + 60 + 4 \\ &= (16 - 48) + (-4 + 60) + 4 \\ &= -32 + 56 + 4 \\ &= 28 \end{align*}$$

\]
Since \( f(-3) = 4 \) and \( f(-2) = 28 \), both are positive, so no zero in \([-3, -2]\).

Step 4: Evaluate \( f(x) \) at \( x = -1 \) and \( x = 0 \)

For \( x = -1 \):
\[

$$\begin{align*} f(-1) &= (-1)^4 + 6(-1)^3 - (-1)^2 - 30(-1) + 4 \\ &= 1 + 6(-1) - 1 + 30 + 4 \\ &= 1 - 6 - 1 + 30 + 4 \\ &= (1 - 6) + (-1 + 30) + 4 \\ &= -5 + 29 + 4 \\ &= 28 \end{align*}$$

\]
For \( x = 0 \):
\[
f(0) = 0^4 + 6(0)^3 - 0^2 - 30(0) + 4 = 4
\]
Both \( f(-1) = 28 \) and \( f(0) = 4 \) are positive, so no zero in \([-1, 0]\).

Step 5: Evaluate \( f(x) \) at \( x = 0 \) and \( x = 1 \)

We know \( f(0) = 4 \).
For \( x = 1 \):
\[

$$\begin{align*} f(1) &= 1^4 + 6(1)^3 - 1^2 - 30(1) + 4 \\ &= 1 + 6 - 1 - 30 + 4 \\ &= (1 + 6) + (-1 - 30) + 4 \\ &= 7 - 31 + 4 \\ &= -20 \end{align*}$$

\]
Since \( f(0) = 4 \) and \( f(1) = -20 \), they have opposite signs, so there is at least one zero in \([0, 1]\).

Step 6: Evaluate \( f(x) \) at \( x = 1 \) and \( x = 2 \)

We know \( f(1) = -20 \).
For \( x = 2 \):
\[

$$\begin{align*} f(2) &= 2^4 + 6(2)^3 - 2^2 - 30(2) + 4 \\ &= 16 + 6(8) - 4 - 60 + 4 \\ &= 16 + 48 - 4 - 60 + 4 \\ &= (16 + 48) + (-4 - 60) + 4 \\ &= 64 - 64 + 4 \\ &= 4 \end{align*}$$

\]
Since \( f(1) = -20 \) and \( f(2) = 4 \), they have opposite signs, so there is at least one zero in \([1, 2]\).

Answer:

The intervals that contain at least one zero are:

• \([-5, -4]\)
• \([-4, -3]\)
• \([0, 1]\)
• \([1, 2]\)