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

graph the function ( y=\frac{2 x}{x^{2}-16} ) by identifying the domain and any symmetries, finding the derivatives ( y^{prime} ) and ( y^{prime prime} ), 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. identify where the curve is concave up or concave down. select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. a. the curve is never concave up and is concave down on the interval(s) (type your answer in interval notation. use a comma to separate answers as needed.) b. the curve is concave up on the interval(s) and is concave down on the interval(s) (type your answer in interval notation. use a comma to separate answers as needed.) c. the curve is concave up on the interval(s) and is never concave down. (type your answer in interval notation. use a comma to separate answers as needed.) d. the curve is neither concave up nor concave down.

graph the function ( y=\frac{2 x}{x^{2}-16} ) by identifying the domain and any symmetries, finding the derivatives ( y^{prime} ) and ( y^{prime prime} ), 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. identify where the curve is concave up or concave down. select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. a. the curve is never concave up and is concave down on the interval(s) (type your answer in interval notation. use a comma to separate answers as needed.) b. the curve is concave up on the interval(s) and is concave down on the interval(s) (type your answer in interval notation. use a comma to separate answers as needed.) c. the curve is concave up on the interval(s) and is never concave down. (type your answer in interval notation. use a comma to separate answers as needed.) d. the curve is neither concave up nor concave down.

Answer

Explanation:

Step1: Find the second - derivative (y'')

First, use the quotient rule (y=\frac{u}{v}), where (u = 2x) and (v=x^{2}-16). The first - derivative (y'=\frac{u'v - uv'}{v^{2}}), with (u'=2) and (v' = 2x). [ \begin{align*} y'&=\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}} \end{align*} ] Now, find the second - derivative (y'') using the quotient rule again ((u=-2(x^{2}+16)), (u'=-4x), (v=(x^{2}-16)^{2}), (v' = 2(x^{2}-16)(2x))) [ \begin{align*} y''&=\frac{-4x(x^{2}-16)^{2}-(-2(x^{2}+16))\times2(x^{2}-16)(2x)}{(x^{2}-16)^{4}}\ &=\frac{-4x(x^{2}-16)+8x(x^{2}+16)}{(x^{2}-16)^{3}}\ &=\frac{-4x^{3}+64x + 8x^{3}+128x}{(x^{2}-16)^{3}}\ &=\frac{4x^{3}+192x}{(x^{2}-16)^{3}}\ &=\frac{4x(x^{2}+48)}{(x - 4)^{3}(x + 4)^{3}} \end{align*} ]

Step2: Find the intervals of concavity

Set (y'' = 0), then (4x(x^{2}+48)=0) (since (x^{2}+48>0) for all real (x)), so (x = 0). The domain of (y) is (x\neq\pm4).

  • Test the intervals ((-\infty,-4)), ((-4,0)), ((0,4)), ((4,\infty))
    • For (x=-5) (in ((-\infty,-4))): (y''=\frac{4\times(-5)\times((-5)^{2}+48)}{((-5)-4)^{3}((-5)+4)^{3}}=\frac{-20\times73}{(-9)^{3}(-1)^{3}}<0)
    • For (x=-1) (in ((-4,0))): (y''=\frac{4\times(-1)\times((-1)^{2}+48)}{((-1)-4)^{3}((-1)+4)^{3}}=\frac{-4\times49}{(-5)^{3}(3)^{3}}>0)
    • For (x = 1) (in ((0,4))): (y''=\frac{4\times1\times(1^{2}+48)}{(1 - 4)^{3}(1 + 4)^{3}}=\frac{4\times49}{(-3)^{3}(5)^{3}}<0)
    • For (x = 5) (in ((4,\infty))): (y''=\frac{4\times5\times(5^{2}+48)}{(5 - 4)^{3}(5 + 4)^{3}}=\frac{20\times73}{(1)^{3}(9)^{3}}>0)

Answer:

B. The curve is concave up on the interval(s) ((-4,0),(4,\infty)) and is concave down on the interval(s) ((-\infty,-4),(0,4))