Summarize the pertinent information oblainec by applying the graphing stralegy and sketch the graph of \( y=f(x) \). \[ f(x)=\left(x^{2}+10\right)\left(100-x^{2}\right) \] Select the correct choice below and, if necessary, fil in the answer box to complete your choice. A. The function 1 has a local maximum a: \( x-3 \sqrt{5}, 3 \sqrt{5} \). (Type an exact answer, using radicals as needed. Use a comma lo separate answers as needed.) B. The function f has no local maximum. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The function \( f \) has a local minimum at \( \mathbf{x}=\mathbf{0} \). (Type an exact answer, using radicals as noeded. Use a comma to separale answers as needed.) B. The function thas no local minimurn. Select the correct choice below and, if necessary, fill in the answer box to corriplete your choice. A. The function I is concave upward on the subinterval(s) \( \square \) . (Type your answer in interval notation Type an exact answer, using ladicals as needed. Use a comma to separale answers as needer B. The function \( f \) is never concave upward.

We begin with the function \[ f(x)=\left(x^2+10\right)(100-x^2). \] It is convenient to expand and simplify: \[ f(x)=x^2(100-x^2)+10(100-x^2)=100x^2-x^4+1000-10x^2=-x^4+90x^2+1000. \] ### Step 1. Find Critical Points First, compute the first derivative: \[ f'(x)=\frac{d}{dx}(-x^4+90x^2+1000)=-4x^3+180x. \] Factor common terms: \[ f'(x)=-4x\left(x^2-45\right). \] Set \( f'(x)=0 \): \[ -4x\left(x^2-45\right)=0. \] Thus, the solutions are: \[ x=0 \quad \text{or} \quad x^2=45 \quad \Rightarrow \quad x=\pm 3\sqrt{5}. \] ### Step 2. Determine Local Maximum and Minimum Next, compute the second derivative: \[ f''(x)=\frac{d}{dx}\left(-4x^3+180x\right)=-12x^2+180=180-12x^2. \] **At \( x=0 \):** \[ f''(0)=180>0, \] so by the second derivative test, \( x=0 \) is a local minimum. **At \( x=3\sqrt{5} \):** \[ f''(3\sqrt{5})=180-12(45)=180-540=-360<0, \] so \( x=3\sqrt{5} \) is a local maximum. Similarly, for \( x=-3\sqrt{5} \): \[ f''(-3\sqrt{5})=180-12(45)=-360<0, \] thus \( x=-3\sqrt{5} \) is a local maximum. **Answer Choices:** - For the local maximum, select **A**: The function has a local maximum at \( x=-3\sqrt{5},\,3\sqrt{5} \). - For the local minimum, select **A**: The function \( f \) has a local minimum at \( x=0 \). ### Step 3. Determine the Concavity The function is concave upward when \( f''(x) > 0 \): \[ 180-12x^2>0 \quad \Longrightarrow \quad 12x^2 < 180 \quad \Longrightarrow \quad x^2<15. \] Thus, the function is concave upward on the interval: \[ (-\sqrt{15},\,\sqrt{15}). \] So for the concavity, select **A**: The function \( f \) is concave upward on the subinterval \( (-\sqrt{15},\,\sqrt{15}) \).