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Find the intervals on which \( f(x) \) is increasing and the intervals on which \( f(x) \) is decreasing. Then sketch the graph. Add horizontal tangent lines. \( f(x)=x^{4}-50 x^{2} \) Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The function is increasing on (Type your answer using interval notation. Use a comma to separate answers as needed.) B. The function is never increasing.

Ask by Barber Romero. in the United States
Mar 11,2025

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The function \( f(x) = x^4 - 50x^2 \) is increasing on the intervals \( (-5, 0) \) and \( (5, \infty) \), and decreasing on \( (-\infty, -5) \) and \( (0, 5) \).

Solución

To determine the intervals on which the function \( f(x) = x^4 - 50x^2 \) is increasing or decreasing, we need to follow these steps: 1. **Find the derivative \( f'(x) \)** to determine the critical points. 2. **Set the derivative equal to zero** to find the critical points. 3. **Analyze the sign of the derivative** in the intervals defined by the critical points to determine where the function is increasing or decreasing. ### Step 1: Find the derivative The derivative of \( f(x) \) is calculated as follows: \[ f'(x) = \frac{d}{dx}(x^4 - 50x^2) = 4x^3 - 100x \] ### Step 2: Set the derivative equal to zero Now, we set the derivative equal to zero to find the critical points: \[ 4x^3 - 100x = 0 \] Factoring out \( 4x \): \[ 4x(x^2 - 25) = 0 \] This gives us: \[ 4x = 0 \quad \text{or} \quad x^2 - 25 = 0 \] From \( 4x = 0 \), we get \( x = 0 \). From \( x^2 - 25 = 0 \), we get \( x = 5 \) and \( x = -5 \). Thus, the critical points are \( x = -5, 0, 5 \). ### Step 3: Analyze the sign of the derivative We will test the intervals defined by the critical points: \( (-\infty, -5) \), \( (-5, 0) \), \( (0, 5) \), and \( (5, \infty) \). 1. **Interval \( (-\infty, -5) \)**: Choose \( x = -6 \) \[ f'(-6) = 4(-6)^3 - 100(-6) = -864 + 600 = -264 \quad (\text{decreasing}) \] 2. **Interval \( (-5, 0) \)**: Choose \( x = -1 \) \[ f'(-1) = 4(-1)^3 - 100(-1) = -4 + 100 = 96 \quad (\text{increasing}) \] 3. **Interval \( (0, 5) \)**: Choose \( x = 1 \) \[ f'(1) = 4(1)^3 - 100(1) = 4 - 100 = -96 \quad (\text{decreasing}) \] 4. **Interval \( (5, \infty) \)**: Choose \( x = 6 \) \[ f'(6) = 4(6)^3 - 100(6) = 864 - 600 = 264 \quad (\text{increasing}) \] ### Summary of Intervals - The function is **decreasing** on \( (-\infty, -5) \) and \( (0, 5) \). - The function is **increasing** on \( (-5, 0) \) and \( (5, \infty) \). ### Final Answer A. The function is increasing on \( (-5, 0) \) and \( (5, \infty) \). Now, let's sketch the graph of \( f(x) \) and add horizontal tangent lines at the critical points. ### Sketching the Graph 1. **Critical Points**: \( f(-5) \), \( f(0) \), and \( f(5) \). 2. **Horizontal Tangent Lines**: At \( x = -5, 0, 5 \). Let's calculate the function values at the critical points: \[ f(-5) = (-5)^4 - 50(-5)^2 = 625 - 1250 = -625 \] \[ f(0) = 0^4 - 50(0)^2 = 0 \] \[ f(5) = 5^4 - 50(5)^2 = 625 - 1250 = -625 \] Now, I will plot the graph and add horizontal tangent lines at these points. ### Graphing the Function The graph will show the behavior of the function with the identified intervals of increase and decrease, along with the horizontal tangent lines at the critical points. Let's proceed to sketch the graph. (Note: As a text-based AI, I cannot create visual graphs directly, but you can use graphing software or tools to plot the function \( f(x) = x^4 - 50x^2 \) and mark the points and tangent lines accordingly.)

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To determine where \( f(x) = x^4 - 50x^2 \) is increasing or decreasing, we first need to find the derivative, \( f'(x) \). Calculating the derivative: \[ f'(x) = 4x^3 - 100x \] Next, we set the derivative to zero to find the critical points: \[ 4x^3 - 100x = 0 \] Factoring out \( 4x \): \[ 4x(x^2 - 25) = 0 \] This gives us: \[ x = 0,\, x = 5,\, x = -5 \] Now, we test intervals around these points: \( (-\infty, -5) \), \( (-5, 0) \), \( (0, 5) \), and \( (5, \infty) \). 1. **For \( x \in (-\infty, -5) \)**, choose \( x = -6 \): \[ f'(-6) = 4(-6)^3 - 100(-6) = -864 + 600 = -264 \quad (\text{decreasing}) \] 2. **For \( x \in (-5, 0) \)**, choose \( x = -1 \): \[ f'(-1) = 4(-1)^3 - 100(-1) = -4 + 100 = 96 \quad (\text{increasing}) \] 3. **For \( x \in (0, 5) \)**, choose \( x = 1 \): \[ f'(1) = 4(1)^3 - 100(1) = 4 - 100 = -96 \quad (\text{decreasing}) \] 4. **For \( x \in (5, \infty) \)**, choose \( x = 6 \): \[ f'(6) = 4(6)^3 - 100(6) = 864 - 600 = 264 \quad (\text{increasing}) \] From our tests, we conclude: - The function is increasing on the intervals \( (-5, 0) \) and \( (5, \infty) \). - The function is decreasing on the intervals \( (-\infty, -5) \) and \( (0, 5) \). Now, filling in the blanks: A. The function is increasing on: **(-5, 0), (5, ∞)**. B. The function is never increasing: **False**. **Horizontal tangents** occur at the critical points where \( f'(x) = 0 \), which are \( x = -5, 0, 5 \). The overall conclusion is to sketch a graph where the curve dips downwards as \( x \) approaches -5, rises as it passes through \( (-5, 0) \), dips again until \( x=5 \), and finally rises towards infinity after \( x=5 \). There are horizontal tangent lines at \( x=-5, 0, 5 \) where the slope is zero.

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