Question: Consider the following polynomial function. \[ f(x) = x^4 + x^3 - 11x^2 - 5x + 30 \] Step 4 of…

Consider the following polynomial function.

\[ f(x) = x^4 + x^3 - 11x^2 - 5x + 30 \]

Step 4 of 4: Find the zero(s) at which \( f \) “flattens out”. Express the zero(s) as ordered pair(s).

Answer

Select the number of zero(s) at which \( f \) “flattens out”.

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Solution

To find the zero(s) at which \( f(x) \) “flattens out,” we need to determine the points where the function \( f(x) \) equals zero and its derivative \( f’(x) \) also equals zero. \[ f(x) = x^4 + x^3 - 11x^2 - 5x + 30 \] \[ f’(x) = 4x^3 + 3x^2 - 22x - 5 \] First, let’s factorize \( f(x) \): \[ f(x) = (x - 2)(x + 3)(x^2 - 5) \] This gives us the roots: \[ x = 2, \quad x = -3, \quad x = \sqrt{5}, \quad x = -\sqrt{5} \] Next, we check if any of these roots satisfy \( f’(x) = 0 \): \[ f’(2) = 4(2)^3 + 3(2)^2 - 22(2) - 5 = 32 + 12 - 44 - 5 = -5 \neq 0 \] \[ f’(-3) = 4(-3)^3 + 3(-3)^2 - 22(-3) - 5 = -108 + 27 + 66 - 5 = -20 \neq 0 \] \[ f’(\sqrt{5}) = 4(\sqrt{5})^3 + 3(\sqrt{5})^2 - 22(\sqrt{5}) - 5 \approx -2\sqrt{5} + 10 \neq 0 \] \[ f’(-\sqrt{5}) = 4(-\sqrt{5})^3 + 3(-\sqrt{5})^2 - 22(-\sqrt{5}) - 5 \approx 2\sqrt{5} + 10 \neq 0 \] None of the roots satisfy \( f’(x) = 0 \). Final Answer: There are no zero(s) at which \( f(x) \) “flattens out.”