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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.

(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

1 answer

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:

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