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Find the intervals on which \( f(x) \) is increasing and the intervals \( f(x)=x^{4}-8 x^{2} \) Select the correct choice below and, if nocessary, fill in the ans A. The function is increasing on (Type your answer using interval notation. Use a comm B. The function is never increasing. Select the correct choice below and, if necessary, fill in the ans A. The function is decreasing on (Type your answer using interval notation. Use a comm B. The function is never decreasing. Choose the corlect graph of \( f(x)=x^{4}-8 x^{2} \) with the correct hori: C

Ask by Henry Fernandez. in the United States
Mar 10,2025

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The function \( f(x) = x^4 - 8x^2 \) is: - **Increasing** on the intervals \((-2, 0)\) and \((2, \infty)\). - **Decreasing** on the intervals \((-\infty, -2)\) and \((0, 2)\). For the graph of \( f(x) = x^4 - 8x^2 \), it should show: - A local maximum at \( x = -2 \), - A local minimum at \( x = 0 \), - A local maximum at \( x = 2 \).

Solución

1. Compute the derivative of the function: \[ f(x)=x^4-8x^2 \] Differentiate term‐by-term: \[ f'(x)=4x^3-16x \] 2. Factor the derivative: \[ f'(x)=4x(x^2-4)=4x(x-2)(x+2) \] The critical points occur when: \[ 4x(x-2)(x+2)=0 \quad \Longrightarrow \quad x=-2,\; x=0,\; x=2. \] 3. Test the sign of \( f'(x) \) on the intervals determined by the critical points: - **Interval \((-\infty,-2)\):** Choose \(x=-3\): \[ f'(-3)=4(-3)((-3)-2)((-3)+2)=4(-3)(-5)(-1)=-60 \quad (\text{negative}) \] So, \(f(x)\) is decreasing on \((-\infty,-2)\). - **Interval \((-2,0)\):** Choose \(x=-1\): \[ f'(-1)=4(-1)((-1)-2)((-1)+2)=4(-1)(-3)(1)=12 \quad (\text{positive}) \] So, \(f(x)\) is increasing on \((-2,0)\). - **Interval \((0,2)\):** Choose \(x=1\): \[ f'(1)=4(1)(1-2)(1+2)=4(1)(-1)(3)=-12 \quad (\text{negative}) \] So, \(f(x)\) is decreasing on \((0,2)\). - **Interval \((2,\infty)\):** Choose \(x=3\): \[ f'(3)=4(3)(3-2)(3+2)=4(3)(1)(5)=60 \quad (\text{positive}) \] So, \(f(x)\) is increasing on \((2,\infty)\). 4. Conclusions: - The function is increasing on: \[ (-2,0)\cup (2, \infty) \] - The function is decreasing on: \[ (-\infty,-2)\cup (0,2) \] 5. Regarding the graph of \( f(x)=x^4-8x^2 \): The correct graph should show local minima and local maxima corresponding to the critical points: - A local maximum at \(x=-2\), - A local minimum at \(x=0\), - A local maximum at \(x=2\). Thus, the answers to the questions are: A. The function is increasing on \(\boxed{(-2,0)\cup (2,\infty)}\). A. The function is decreasing on \(\boxed{(-\infty,-2)\cup (0,2)}\).

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Extra Insights

To determine where the function \( f(x) = x^{4} - 8x^{2} \) is increasing or decreasing, we first find its derivative. This gives us \( f'(x) = 4x^{3} - 16x \). Setting the derivative equal to zero to find critical points, we solve \( 4x(x^{2} - 4) = 0 \), leading to \( x = 0, 2, -2 \). Now we can analyze the sign of the derivative in the intervals defined by these critical points: 1. For \( (-\infty, -2) \): Choose \( x = -3 \) → \( f'(-3) > 0 \) (increasing). 2. For \( (-2, 0) \): Choose \( x = -1 \) → \( f'(-1) < 0 \) (decreasing). 3. For \( (0, 2) \): Choose \( x = 1 \) → \( f'(1) < 0 \) (decreasing). 4. For \( (2, \infty) \): Choose \( x = 3 \) → \( f'(3) > 0 \) (increasing). Therefore, the function is increasing on the intervals \( (-\infty, -2) \) and \( (2, \infty) \), and it is decreasing on \( (-2, 0) \) and \( (0, 2) \). --- Picture this: a roller coaster ride! Imagine a function peaking and dipping through the critical points as if they were the ups and downs of the ride. Riders get their thrill from the climbs between -∞ to -2 and 2 to ∞ where \( f \) increases and the more tame but thrilling drops from -2 to 0 and 0 to 2, creating that balance of happiness and exhilaration through the ride of mathematics! For further exploration, you can look into more complex polynomial functions or study derivatives deeper. Checking out calculus textbooks or online platforms that illustrate polynomial behavior can further enhance your understanding! Plus, there are tons of videos out there explaining derivatives in a fun way.

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