Questions: (D) f is a polynomial of degree 3 with a leading coefficient of 9. Values of the polynomial function f for selected values of x are given in the table. x -1 1 3 5 7 f(x) -36 0 4 0 12 If all of the zeros of the function f are given in the table, which of the following must be true? (A) The function f has a local minimum at (-1,-36). (B) The function f has a local minimum at (5,0). (C) The function f has a local maximum at (3,4). (D) The function f has a local maximum at (1,0).

Solution Steps

To determine which of the given statements about the polynomial function \( f \) are true, we need to analyze the given data points and the nature of the polynomial. Since \( f \) is a cubic polynomial with a leading coefficient of 9, we can use the given points to infer the behavior of the function.

1. Identify the zeros of the polynomial from the table.
2. Use the zeros to determine the intervals where the function changes sign.
3. Analyze the function's behavior around the zeros to determine local minima and maxima.

Given the polynomial \( f(x) = 0.5x^3 - 5.5x^2 + 17.5x - 12.5 \), we first find its derivative: \[ f'(x) = 1.5x^2 - 11x + 17.5 \]

To find the critical points, we solve \( f'(x) = 0 \): \[ 1.5x^2 - 11x + 17.5 = 0 \] The solutions are: \[ x = 5 \] \[ x \approx 2.3333 \]

We evaluate the second derivative \( f''(x) = 3x - 11 \) at the critical points: \[ f''(5) = 3(5) - 11 = 4 \] \[ f''(2.3333) \approx 3(2.3333) - 11 \approx -4 \]

Since \( f''(5) > 0 \), \( x = 5 \) is a local minimum. Since \( f''(2.3333) < 0 \), \( x \approx 2.3333 \) is a local maximum.

• (A) The function \( f \) has a local minimum at \((-1, -36)\). This is incorrect because the local minimum is at \( x = 5 \).
• (B) The function \( f \) has a local minimum at \((5, 0)\). This is incorrect because the local minimum is at \( x = 5 \), but \( f(5) = 0 \).
• (C) The function \( f \) has a local maximum at \((3, 4)\). This is incorrect because the local maximum is at \( x \approx 2.3333 \).
• (D) The function \( f \) has a local maximum at \((1, 0)\). This is incorrect because \( x = 1 \) is a zero, not a local maximum.

Final Answer