Summaree the pertinent information oblained by applying the graphing slralogy ard skotch the graph of $$ y=I(x) $$. $$ f(x)=\left(x^{2}+10\right)\left(100-x^{2}\right) $$ Solect the correct choice bolow and, if necessary, fill in the anewor box to complate your choico. A. The $$ y $$-inlercepl of fis $$ y=1000 $$. (Type an exacl answer, using radicals as needed) B. The function f has no y -inforcool. Seleci the correct choice below and, if necussary, fill in the answer box to complate your choica. A. The function fis increasing on the subinlerval(s) $$ (-\infty,-3 \sqrt{5}),(0,3 \sqrt{5}) $$. (Type your unswer in interval nulation. Typuan ukiel arower, using raxdicals as neoded. Use a comma lo separale answors as neoded.) B. The furctun it inever increas ing. Soloct the corrocl choice below and, if necessary, fil in the answer box to complote your choire. A. The function fis decroasing on tho subinierval $$ (s) $$ $$ \square $$ . (Type your answet in interval notation. I ypo an oxact answer, Lsing radicals as needod. Use a comma lo separale misweis as needed.) B. The function fis never decreasing.

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1. When $$ x=0 $$, the function gives $$ f(0) = (0^2+10)(100-0^2)=10\cdot 100 = 1000. $$ Thus, the $$ y $$-intercept is $$ 1000 $$. 2. To find the intervals where $$ f $$ is increasing or decreasing, we differentiate $$ f(x)=\bigl(x^2+10\bigr)\bigl(100-x^2\bigr). $$ Differentiating via the product rule: $$ f'(x)=\frac{d}{dx}(x^2+10)\cdot (100-x^2) + (x^2+10)\cdot\frac{d}{dx}(100-x^2). $$ $$ = 2x(100-x^2) + (x^2+10)(-2x). $$ Factor out $$2x$$: $$ f'(x)=2x\Bigl[(100-x^2)-(x^2+10)\Bigr]=2x\Bigl[100-x^2-x^2-10\Bigr]. $$ Simplify inside the brackets: $$ =2x\Bigl[90-2x^2\Bigr]=4x\Bigl[45-x^2\Bigr]. $$ 3. The critical points occur when $$ 4x(45-x^2)=0 $$ which gives $$ x=0 \quad \text{or} \quad 45-x^2=0 \Rightarrow x^2=45 \Rightarrow x=\pm\sqrt{45}=\pm 3\sqrt{5}. $$ 4. Testing intervals determined by $$ x=-3\sqrt{5} $$, $$ 0 $$, and $$ 3\sqrt{5} $$: - For $$ x < -3\sqrt{5} $$: Let $$ x=-10 $$. Then, $$45-x^2<0$$ (since $$x^2>45$$) and $$x$$ is negative. The product $$4x(45-x^2)$$ is positive (a negative times a negative is positive). Thus, $$ f $$ is increasing on $$(-\infty,-3\sqrt{5})$$. - For $$ -3\sqrt{5} < x < 0 $$: $$ x $$ is negative and $$45-x^2>0$$, so $$ f'(x) $$ is negative. Hence, $$ f $$ is decreasing on $$(-3\sqrt{5},0)$$. - For $$ 0 < x < 3\sqrt{5} $$: $$ x $$ is positive and $$45-x^2>0$$, so $$ f'(x) $$ is positive. Thus, $$ f $$ is increasing on $$(0,3\sqrt{5})$$. - For $$ x>3\sqrt{5} $$: $$ x $$ is positive but $$45-x^2<0$$, so $$ f'(x) $$ is negative. Therefore, $$ f $$ is decreasing on $$(3\sqrt{5},\infty)$$. 5. Answers: - $$ y $$-intercept: $$ 1000 $$. - $$ f $$ is increasing on the subintervals: $$ (-\infty,-3\sqrt{5}),\;(0,3\sqrt{5}). $$ - $$ f $$ is decreasing on the subintervals: $$ (-3\sqrt{5},0),\;(3\sqrt{5},\infty). $$ - The $$ y $$-intercept of the function $$ f(x) = (x^2 + 10)(100 - x^2) $$ is $$ 1000 $$. - The function $$ f $$ is increasing on the intervals $$ (-\infty, -3\sqrt{5}) $$ and $$ (0, 3\sqrt{5}) $$. - The function $$ f $$ is decreasing on the intervals $$ (-3\sqrt{5}, 0) $$ and $$ (3\sqrt{5}, \infty) $$.

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