Pregunta
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Find the intervals on which \( f(x) \) is increasing, the intervals on which \( f(x) \) is decreasing, and the local extrema. \( f(x)=-3 x^{2}-30 x-21 \) Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The function is increasing on ( \( -\infty,-5 \) ). (Type your answer in interval notation. Type integers or simplified fractions. Use a comma to separate answers as needed.) The function is never increasing. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The function is decreasing on (Type your answer in interval notation. Type integers or simplified fractions. Use a comma to separate answers as needed.) B. The function is never decreasing.

Ask by Malone Johnson. in the United States
Mar 12,2025

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The function is increasing on \((-\infty, -5)\) and decreasing on \((-5, \infty)\). There is a local maximum at \((-5, 54)\).

Solución

**Step 1. Find the derivative of \( f(x) \).** We have \[ f(x)=-3x^2-30x-21. \] Differentiate with respect to \( x \): \[ f'(x) = \frac{d}{dx}\left(-3x^2-30x-21\right) = -6x-30. \] **Step 2. Find the critical point by setting \( f'(x) = 0 \).** Set \[ -6x-30=0. \] Solve for \( x \): \[ -6x = 30 \quad \Longrightarrow \quad x=-5. \] **Step 3. Determine the intervals on which \( f(x) \) is increasing or decreasing.** Write \( f'(x) \) in factored form: \[ f'(x)=-6(x+5). \] - For \( x < -5 \): The term \( x+5 < 0 \), so \[ -6(x+5) > 0. \] Hence, \( f'(x) > 0 \), and \( f(x) \) is increasing. - For \( x > -5 \): The term \( x+5 > 0 \), so \[ -6(x+5) < 0. \] Hence, \( f'(x) < 0 \), and \( f(x) \) is decreasing. Thus, the function is increasing on \((-\infty,-5)\) and decreasing on \((-5,\infty)\). **Step 4. Identify the local extrema.** Since \( f(x) \) changes from increasing to decreasing at \( x=-5 \), there is a local maximum at \( x = -5 \). Find the local maximum value: \[ f(-5) = -3(-5)^2 - 30(-5) - 21. \] Calculate: \[ f(-5) = -3(25) + 150 - 21 = -75 + 150 - 21 = 54. \] So, there is a local maximum at \((-5, 54)\). **Final Answers:** - The function is increasing on \((-\infty, -5)\). - The function is decreasing on \((-5, \infty)\). - There is a local maximum at \((-5, 54)\).

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Bonus Knowledge

To determine the intervals of increase and decrease for the function \( f(x) = -3x^2 - 30x - 21 \), we first need to find its derivative: \[ f'(x) = -6x - 30 \] Setting the derivative equal to zero to find critical points: \[ -6x - 30 = 0 \implies -6x = 30 \implies x = -5 \] Now, we analyze the sign of the derivative: - For \( x < -5 \), \( f'(-6) = -6(-6) - 30 = 36 - 30 = 6 \) (positive) - For \( x > -5 \), \( f'(-4) = -6(-4) - 30 = 24 - 30 = -6 \) (negative) Thus, the function is increasing on \( (-\infty, -5) \) and decreasing on \( (-5, \infty) \). To find the local extrema, we evaluate \( f(x) \) at the critical point \( x = -5 \): \[ f(-5) = -3(-5^2) - 30(-5) - 21 = -3(25) + 150 - 21 = -75 + 150 - 21 = 54 \] So, there is a local maximum at \( (-5, 54) \). To summarize: A. The function is increasing on \( (-\infty, -5) \). B. The function is decreasing on \( (-5, \infty) \).

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