Find the Concavity f(x)=x/(x^2+1)

Problem

ƒ(x)=x/(x^2+1)

Solution

1. Find the first derivative using the quotient rule d()/d(x)u/v=(vd(u)/d(x)−ud(v)/d(x))/(v^2)

ƒ(x)^′=((x^2+1)*(1)−x*(2*x))/((x^2+1)^2)

ƒ(x)^′=(1−x^2)/((x^2+1)^2)

2. Find the second derivative by applying the quotient rule again to ƒ(x)^′

ƒ(x)^″=((x^2+1)^2*(−2*x)−(1−x^2)*(2*(x^2+1)*(2*x)))/((x^2+1)^4)

3. Simplify the second derivative by factoring out (x^2+1) from the numerator.

ƒ(x)^″=((x^2+1)*[−2*x*(x^2+1)−4*x*(1−x^2)])/((x^2+1)^4)

ƒ(x)^″=(−2*x^3−2*x−4*x+4*x^3)/((x^2+1)^3)

ƒ(x)^″=(2*x^3−6*x)/((x^2+1)^3)

ƒ(x)^″=(2*x*(x^2−3))/((x^2+1)^3)

4. Identify the critical points for concavity by setting ƒ(x)^″=0

2*x*(x^2−3)=0

x=0,x=√(,3),x=−√(,3)

5. Test the intervals created by these points in ƒ(x)^″ to determine the sign.

Interval *(−∞,−√(,3)):ƒ^″*(−2)<0⇒Concave Down

Interval *(−√(,3),0):ƒ^″*(−1)>0⇒Concave Up

Interval *(0,√(,3)):ƒ(1)^″<0⇒Concave Down

Interval *(√(,3),∞):ƒ(2)^″>0⇒Concave Up

Final Answer

Concave Up: *(−√(,3),0)∪(√(,3),∞), Concave Down: *(−∞,−√(,3))∪(0,√(,3))

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