Find the Local Maxima and Minima f(x)=8/(x^2+1)

Problem

ƒ(x)=8/(x2+1)

Solution

1. Find the first derivative of the function using the quotient rule or the chain rule by rewriting the function as ƒ(x)=8*(x2+1)(−1)

d(ƒ(x))/d(x)=−8*(x2+1)(−2)⋅2*x

d(ƒ(x))/d(x)=(−16*x)/((x2+1)2)

2. Identify critical points by setting the first derivative equal to zero and solving for x

(−16*x)/((x2+1)2)=0

−16*x=0

x=0

3. Find the second derivative to apply the Second Derivative Test, using the quotient rule on ƒ(x)′

d2(ƒ(x))/(d(x)2)=(−16*(x2+1)2−(−16*x)*(2*(x2+1)*(2*x)))/((x2+1)4)

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

d2(ƒ(x))/(d(x)2)=(48*x2−16)/((x2+1)3)

4. Evaluate the second derivative at the critical point x=0 to determine the nature of the extremum.

ƒ(0)″=(48*(0)2−16)/((0+1)3)

ƒ(0)″=−16

Since ƒ(0)″<0 a local maximum occurs at x=0

5. Calculate the function value at x=0 to find the coordinates of the local maximum.

ƒ(0)=8/(0+1)

ƒ(0)=8

6. Analyze the behavior of the function as x→±∞ to check for other extrema. Since ƒ(x)→0 as x→±∞ and there are no other critical points, there are no local minima.

Final Answer

Local Maxima: *(0,8), Local Minima: None

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