graph the function ( y=\frac{2x}{x^{2}-16} ) by identifying the domain and any symmetries, finding the…

graph the function ( y=\frac{2x}{x^{2}-16} ) by identifying the domain and any symmetries, finding the derivatives ( y ) and ( y ), finding the critical points and identifying the functions behavior at each one, finding where the curve is increasing and where it is decreasing, finding the points of inflection, determining the concavity of the curve, identifying any asymptotes, and plotting any key points such as intercepts, critical points, and inflection points. then find coordinates of absolute extreme points, if any. \n\na. the absolute maximum value occurs at ( x= ). (use a comma to separate answers as needed. type each answer only once.) \nb. there is no absolute maximum. \n\nidentify the absolute minimum value and where it occurs. select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. \n\na. the absolute minimum value occurs at ( x= ). (use a comma to separate answers as needed. type each answer only once.) \nb. there is no absolute minimum.

graph the function ( y=\frac{2x}{x^{2}-16} ) by identifying the domain and any symmetries, finding the derivatives ( y ) and ( y ), finding the critical points and identifying the functions behavior at each one, finding where the curve is increasing and where it is decreasing, finding the points of inflection, determining the concavity of the curve, identifying any asymptotes, and plotting any key points such as intercepts, critical points, and inflection points. then find coordinates of absolute extreme points, if any. \n\na. the absolute maximum value occurs at ( x= ). (use a comma to separate answers as needed. type each answer only once.) \nb. there is no absolute maximum. \n\nidentify the absolute minimum value and where it occurs. select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. \n\na. the absolute minimum value occurs at ( x= ). (use a comma to separate answers as needed. type each answer only once.) \nb. there is no absolute minimum.

Answer

Explanation:

Step1: Analyze the domain

The function ( y=\frac{2x}{x^{2}-16}=\frac{2x}{(x + 4)(x-4)}). The denominator cannot be zero, so the domain is (x\in(-\infty,-4)\cup(-4,4)\cup(4,\infty)).

Step2: Check for symmetry

Replace (x) with (-x), (y(-x)=\frac{-2x}{x^{2}-16}=-y(x)), so the function is odd (symmetric about the origin).

Step3: Find the first - derivative

Using the quotient rule ((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}), where (u = 2x), (u^\prime=2), (v=x^{2}-16), (v^\prime = 2x). (y^\prime=\frac{2(x^{2}-16)-2x(2x)}{(x^{2}-16)^{2}}=\frac{2x^{2}-32 - 4x^{2}}{(x^{2}-16)^{2}}=\frac{-2x^{2}-32}{(x^{2}-16)^{2}}=\frac{-2(x^{2}+16)}{(x^{2}-16)^{2}}) Since (y^\prime=\frac{-2(x^{2}+16)}{(x^{2}-16)^{2}}<0) for all (x) in the domain ((x^{2}+16>0) and ((x^{2}-16)^{2}>0) for (x\neq\pm4)), the function is decreasing on ((-\infty,-4)), ((-4,4)) and ((4,\infty)). There are no critical points (since (y^\prime\neq0) for all (x) in the domain).

Step4: Find the second - derivative

Using the quotient rule again. Let (u=-2x^{2}-32), (u^\prime=-4x), (v=(x^{2}-16)^{2}), (v^\prime = 2(x^{2}-16)(2x)=4x(x^{2}-16)) (y^{\prime\prime}=\frac{-4x(x^{2}-16)^{2}-(-2x^{2}-32)\times4x(x^{2}-16)}{(x^{2}-16)^{4}}) (=\frac{-4x(x^{2}-16)[(x^{2}-16)+(2x^{2}+32)]}{(x^{2}-16)^{4}}=\frac{-4x(3x^{2}+16)}{(x^{2}-16)^{3}}) Set (y^{\prime\prime}=0), then (x = 0) (since (3x^{2}+16>0) for all (x)). When (x<0,x\in(-\infty,-4)\cup(-4,0)), (y^{\prime\prime}>0) (concave up). When (x>0,x\in(0,4)\cup(4,\infty)), (y^{\prime\prime}<0) (concave down). The inflection point is ((0,0)).

Step5: Find asymptotes

Vertical asymptotes: (x = 4) and (x=-4) (since (\lim_{x\rightarrow4^{\pm}}\frac{2x}{x^{2}-16}=\pm\infty) and (\lim_{x\rightarrow - 4^{\pm}}\frac{2x}{x^{2}-16}=\mp\infty)) Horizontal asymptote: (\lim_{x\rightarrow\pm\infty}\frac{2x}{x^{2}-16}=\lim_{x\rightarrow\pm\infty}\frac{\frac{2x}{x^{2}}}{\frac{x^{2}}{x^{2}}-\frac{16}{x^{2}}}=0) (so (y = 0)).

Step6: Analyze absolute extrema

Since the function is decreasing on ((-\infty,-4)), ((-4,4)) and ((4,\infty)) and the domain is not a closed - and - bounded interval (it has vertical asymptotes and extends to (\pm\infty)), there are no absolute maximum and no absolute minimum values.

Answer:

B. There is no absolute maximum. B. There is no absolute minimum.