Question

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.

a. the local minimum/minima is/are located at
(type an ordered pair. use a comma to separate answers as needed.)
b. there are no local minima.
identify any local maxima. select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the local maximum/maxima is/are located at
(type an ordered pair. use a comma to separate answers as needed.)
b. there are no local maxima.

Explanation:

Step1: Find the first derivative \(y'\)

Use the quotient rule \((\frac{u}{v})'=\frac{u'v - uv'}{v^{2}}\), where \(u = 2x\), \(u'=2\), \(v=x^{2}-16\), \(v' = 2x\).

$$
LATEXBLOCK0
$$

Since \(x^{2}+16>0\) and \((x^{2}-16)^{2}>0\) for all \(x
eq\pm4\), \(y'<0\) for all \(x
eq\pm4\).

Step2: Analyze critical points

Critical points occur where \(y' = 0\) or \(y'\) is undefined. \(y'\) is never \(0\) (because \(-2(x^{2}+16)
eq0\) for all real \(x\)), and \(y'\) is undefined at \(x = 4\) and \(x=-4\) (but these are not in the domain of the original function \(y=\frac{2x}{x^{2}-16}\), domain is \(x\in(-\infty,-4)\cup(-4,4)\cup(4,\infty)\)).

Since the function is always decreasing (because \(y'<0\) on its domain), there are no local maxima or minima.

Answer:

B. There are no local minima.
B. There are no local maxima.