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 critical point(s) occur(s) at ( x= ).\n(use a comma to separate answers as needed.)\nb. there are no critical points.\n\nidentify any local minima. select the correct choice below and, if necessary, fill in the answer box to complete your choice.\n\na. the local minimum/minima is/are located at.\n(type an ordered pair. use a comma to separate answers as needed.)\nb. there are no local minima.

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 critical point(s) occur(s) at ( x= ).\n(use a comma to separate answers as needed.)\nb. there are no critical points.\n\nidentify any local minima. select the correct choice below and, if necessary, fill in the answer box to complete your choice.\n\na. the local minimum/minima is/are located at.\n(type an ordered pair. use a comma to separate answers as needed.)\nb. there are no local minima.

Answer

Explanation:

Step1: Find the first derivative (y')

Use the quotient rule ((\frac{u}{v})'=\frac{u'v - uv'}{v^{2}}), where (u = 2x), (u'=2), (v=x^{2}-16), (v' = 2x). [ \begin{align*} y'&=\frac{2(x^{2}-16)-2x\times(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}} \end{align*} ] Set (y' = 0). Since (x^{2}+16>0) for all real (x), the numerator (-2(x^{2}+16)\neq0) for all real (x).

Step2: Analyze critical points

Critical points occur where (y' = 0) or (y') is undefined. The derivative (y') is undefined when (x^{2}-16=0), i.e., (x = 4) or (x=- 4). But at (x = 4) and (x=-4), the function (y=\frac{2x}{x^{2}-16}) is also undefined (since the denominator is zero). So, there are no critical points where the function is defined.

Step3: Analyze local minima

Since there are no critical points (points in the domain of the function where (y'=0) or (y') is undefined and the function is defined), there are no local minima.

Answer:

B. There are no critical points. B. There are no local minima.