Find the Inflection Points f(x)=x/(4x^2-1)

Problem

ƒ(x)=x/(4*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)

d(ƒ(x))/d(x)=((4*x2−1)*(1)−x*(8*x))/((4*x2−1)2)

d(ƒ(x))/d(x)=(−4*x2−1)/((4*x2−1)2)

2. Find the second derivative by applying the quotient rule again to the first derivative.

d2(ƒ(x))/(d(x)2)=((4*x2−1)2*(−8*x)−(−4*x2−1)*(2*(4*x2−1)*(8*x)))/((4*x2−1)4)

3. Simplify the second derivative by factoring out (4*x2−1) from the numerator and canceling it with one factor in the denominator.

d2(ƒ(x))/(d(x)2)=((4*x2−1)*[−8*x*(4*x2−1)−16*x*(−4*x2−1)])/((4*x2−1)4)

d2(ƒ(x))/(d(x)2)=(−32*x3+8*x+64*x3+16*x)/((4*x2−1)3)

d2(ƒ(x))/(d(x)2)=(32*x3+24*x)/((4*x2−1)3)

d2(ƒ(x))/(d(x)2)=(8*x*(4*x2+3))/((4*x2−1)3)

4. Identify potential inflection points by setting the second derivative equal to zero.

(8*x*(4*x2+3))/((4*x2−1)3)=0

8*x*(4*x2+3)=0

x=0

5. Check for concavity changes around x=0 and vertical asymptotes x=±0.5 Since 4*x2+3 is always positive, the sign of ƒ(x)″ depends on x and the denominator. At x=0 the sign of ƒ(x)″ changes from positive to negative, confirming an inflection point.

6. Calculate the y-coordinate for the inflection point by substituting x=0 into the original function.

ƒ(0)=0/(4*(0)2−1)=0

Final Answer

Inflection Point: *(0,0)

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