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

Problem

ƒ(x)=x/(x2+1)

Solution

1. Find the first derivative using the quotient rule d()/d(x)u/v=(vd(u)/d(x)−ud(v)/d(x))/(v2)

ƒ(x)′=((x2+1)*(1)−x*(2*x))/((x2+1)2)

ƒ(x)′=(1−x2)/((x2+1)2)

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

ƒ(x)″=((x2+1)2*(−2*x)−(1−x2)*(2*(x2+1)*(2*x)))/((x2+1)4)

3. Simplify the second derivative by factoring out (x2+1) from the numerator.

ƒ(x)″=((x2+1)*[−2*x*(x2+1)−4*x*(1−x2)])/((x2+1)4)

ƒ(x)″=(−2*x3−2*x−4*x+4*x3)/((x2+1)3)

ƒ(x)″=(2*x3−6*x)/((x2+1)3)

ƒ(x)″=(2*x*(x2−3))/((x2+1)3)

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

2*x*(x2−3)=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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