Find the Inflection Points (x^2)/(x^2+3)

Problem

ƒ(x)=(x^2)/(x^2+3)

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+3)*(2*x)−(x^2)*(2*x))/((x^2+3)^2)

ƒ(x)^′=(2*x^3+6*x−2*x^3)/((x^2+3)^2)

ƒ(x)^′=(6*x)/((x^2+3)^2)

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

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

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

ƒ(x)^″=((x^2+3)*[6*(x^2+3)−24*x^2])/((x^2+3)^4)

ƒ(x)^″=(6*x^2+18−24*x^2)/((x^2+3)^3)

ƒ(x)^″=(18−18*x^2)/((x^2+3)^3)

4. Set the second derivative to zero to find potential inflection points.

(18*(1−x^2))/((x^2+3)^3)=0

1−x^2=0

x^2=1

x=1,x=−1

5. Verify the concavity change by checking the sign of ƒ(x)^″ around the critical values. Since the denominator (x^2+3)^3 is always positive, the sign is determined by 18*(1−x^2) The sign changes at x=−1 and x=1 confirming they are inflection points.

6. Calculate the y-coordinates by substituting the x values back into the original function ƒ(x)

ƒ(1)=1^2/(1^2+3)=1/4

ƒ*(−1)=((−1)^2)/((−1)^2+3)=1/4

Final Answer

Inflection Points=(1,1/4),(−1,1/4)

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