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

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 \(\boxed{(-\infty,-2)\cup (0,2)}\).