Summarize the pertinent information oblained by applying the graphing strategy and skotch 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, fill in the answer box lo comp'ete your chcice. A. The function I is increasing on the subinterval(s) \( (-\infty,-3 \sqrt{5}),(0,3 \sqrt{5}) \). (Type your answer in interval notation. Type an exact answer, using radicals as needed. Use a comma lo separate answers as needed.) B. The function \( f \) is never increasing. Select the corred choice helow and, if necessary, fill in tho answer box to complete your choce. A. The function \( f \) is decreasing on the subinterval(s) \( (-3 \sqrt{5}, 0),(3 \sqrt{5}, \infty) \). (Type your answer in interval notation. Type an oxact answer, using radicals as needed. Use a comma to separate answers as needed ) B. The function I is never decreasing. Select the cortect choice below and, if necessary, fill in the answer box to complete your choice. A. The function f has a local makimum at \( \mathrm{x}= \) \( \square \) (Iype an exact answer, using radicals as needed. Use a comma to soparalo answers as needed.) B. The function f has no local maximum

We begin with \[ f(x) = \left(x^2+10\right)\left(100-x^2\right). \] It is helpful to first expand the function: \[ f(x) = (x^2+10)(100-x^2)=100x^2+1000-x^4-10x^2 = -x^4+90x^2+1000. \] Because the highest‐degree term is \(-x^4\), the graph will eventually go down (i.e. the left and right ends go to \(-\infty\)). Notice also that \( f \) is even, so the graph is symmetric about the \( y \)-axis. ### Step 1. Find the Derivative Differentiate \( f(x) \) to find where the function is increasing or decreasing: \[ f'(x)=\frac{d}{dx}\Bigl[-x^4+90x^2+1000\Bigr] = -4x^3+180x. \] Factor the derivative: \[ f'(x) = -4x^3+180x = -4x(x^2-45). \] Set \( f'(x)=0 \) to get the critical points: \[ -4x(x^2-45)=0. \] Thus, \[ x=0 \quad \text{or} \quad x^2 = 45 \quad \Longrightarrow \quad x=\pm\sqrt{45} = \pm 3\sqrt{5}. \] ### Step 2. Analyze the Sign of \( f'(x) \) Divide the real line into four intervals based on the critical points: \[ (-\infty,-3\sqrt{5}),\quad (-3\sqrt{5},0),\quad (0,3\sqrt{5}),\quad (3\sqrt{5},\infty). \] 1. **For \( x<-3\sqrt{5} \):** Choose \( x=-10 \) (note that \( -10<-3\sqrt{5} \) since \( 3\sqrt{5}\approx 6.7 \)). Then, \[ f'(x)=-4(-10)\Bigl((-10)^2-45\Bigr) = 40(100-45)=40(55)>0. \] So, \( f \) is increasing on \((-\infty,-3\sqrt{5})\). 2. **For \( -3\sqrt{5}0. \] So, \( f \) is increasing on \((0,3\sqrt{5})\). 4. **For \( x>3\sqrt{5} \):** Choose \( x=10 \). Then, \[ f'(x)=-4(10)\Bigl(100-45\Bigr)=-40(55)=-2200<0. \] So, \( f \) is decreasing on \((3\sqrt{5},\infty)\). ### Step 3. Determine the Local Extrema - At \( x = -3\sqrt{5} \): \( f'(x) \) changes from positive to negative. Thus, \( f \) has a local maximum at \( x=-3\sqrt{5} \). - At \( x = 0 \): \( f'(x) \) changes from negative (to the left) to positive (to the right). Thus, \( f \) has a local minimum at \( x=0 \). - At \( x = 3\sqrt{5} \): \( f'(x) \) changes from positive to negative. Thus, \( f \) has a local maximum at \( x=3\sqrt{5} \). Because \( f \) is an even function, the local maximum values at \( x=-3\sqrt{5} \) and \( x=3\sqrt{5} \) are equal (in fact, both are global maxima on their respective sides). ### Final Answers 1. The function \( f \) is increasing on the subintervals \[ (-\infty,-3\sqrt{5}) \quad \text{and} \quad (0,3\sqrt{5}). \] (Correct choice: A) 2. The function \( f \) is decreasing on the subintervals \[ (-3\sqrt{5},0) \quad \text{and} \quad (3\sqrt{5},\infty). \] (Correct choice: A) 3. The function \( f \) has a local maximum at \[ x=-3\sqrt{5} \quad \text{and} \quad x=3\sqrt{5}. \] (Correct choice: A)