Find the Concavity f(x)=(x^2)/(x^2+3)

Problem

ƒ(x)=(x2)/(x2+3)

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+3)*(2*x)−(x2)*(2*x))/((x2+3)2)

ƒ(x)′=(2*x3+6*x−2*x3)/((x2+3)2)

ƒ(x)′=(6*x)/((x2+3)2)

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

ƒ(x)″=((x2+3)2*(6)−(6*x)*(2*(x2+3)*(2*x)))/((x2+3)4)

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

ƒ(x)″=((x2+3)*[6*(x2+3)−24*x2])/((x2+3)4)

ƒ(x)″=(6*x2+18−24*x2)/((x2+3)3)

ƒ(x)″=(18−18*x2)/((x2+3)3)

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

18−18*x2=0

x2=1

x=1,x=−1

5. Test intervals to determine the sign of ƒ(x)″ The denominator (x2+3)3 is always positive.
   For (−∞,−1) let x=−2 ƒ″*(−2)=(18−72)/(p*o*s())<0 (Concave Down).
   For (−1,1) let x=0 ƒ(0)″=(18−0)/(p*o*s())>0 (Concave Up).
   For (1,∞) let x=2 ƒ(2)″=(18−72)/(p*o*s())<0 (Concave Down).

Final Answer

Concave Up: *(−1,1), Concave Down: *(−∞,−1)∪(1,∞)

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