Find the Inflection Points (e^x)/(8+e^x)

Problem

ƒ(x)=(e^x)/(8+e^x)

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)^′=((8+e^x)*e^x−e(e^x)^x)/((8+e^x)^2)

2. Simplify the first derivative by distributing and combining terms in the numerator.

ƒ(x)^′=(8*e^x+e^(2*x)−e^(2*x))/((8+e^x)^2)

ƒ(x)^′=(8*e^x)/((8+e^x)^2)

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

ƒ(x)^″=((8+e^x)^2*(8*e^x)−8*e^x⋅2*(8+e^x)*e^x)/((8+e^x)^4)

4. Factor and simplify the second derivative to find the points where ƒ(x)^″=0

ƒ(x)^″=(8*e^x*(8+e^x)*[(8+e^x)−2*e^x])/((8+e^x)^4)

ƒ(x)^″=(8*e^x*(8−e^x))/((8+e^x)^3)

5. Solve for x by setting the numerator of the second derivative equal to zero.

8−e^x=0

e^x=8

x=ln(8)

6. Calculate the y-coordinate by substituting x=ln(8) back into the original function ƒ(x)

ƒ*(ln(8))=(e^ln(8))/(8+e^ln(8))

ƒ*(ln(8))=8/(8+8)

ƒ*(ln(8))=1/2

7. Verify the inflection point by checking for a sign change in ƒ(x)^″ around x=ln(8) Since 8−e^x changes from positive to negative, an inflection point exists.

Final Answer

Inflection Point=(ln(8),1/2)

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