Find the Inflection Points f(x)=1/(x^2+7)

Problem

ƒ(x)=1/(x2+7)

Solution

1. Find the first derivative by applying the power rule or quotient rule to ƒ(x)=(x2+7)(−1)

d(ƒ(x))/d(x)=−(x2+7)(−2)⋅2*x

d(ƒ(x))/d(x)=(−2*x)/((x2+7)2)

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

d2(ƒ(x))/(d(x)2)=(−2*(x2+7)2−(−2*x)⋅2*(x2+7)⋅2*x)/((x2+7)4)

3. Simplify the second derivative by factoring out (x2+7) from the numerator.

d2(ƒ(x))/(d(x)2)=((x2+7)*(−2*(x2+7)+8*x2))/((x2+7)4)

d2(ƒ(x))/(d(x)2)=(−2*x2−14+8*x2)/((x2+7)3)

d2(ƒ(x))/(d(x)2)=(6*x2−14)/((x2+7)3)

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

(6*x2−14)/((x2+7)3)=0

6*x2−14=0

5. Solve for x by isolating the variable.

6*x2=14

x2=7/3

x=±√(,7/3)

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

ƒ*(±√(,7/3))=1/((±√(,7/3))2+7)

ƒ*(±√(,7/3))=1/(7/3+7)

ƒ*(±√(,7/3))=1/28/3

ƒ*(±√(,7/3))=3/28

Final Answer

Inflection Points=(√(,7/3),3/28),(−√(,7/3),3/28)

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