Find the Local Maxima and Minima f(x)=x/(x^2+9)

Problem

ƒ(x)=x/(x2+9)

Solution

1. Find the derivative using the quotient rule, where u=x and v=x2+9

d(ƒ(x))/d(x)=((1)*(x2+9)−(x)*(2*x))/((x2+9)2)

2. Simplify the derivative by combining like terms in the numerator.

d(ƒ(x))/d(x)=(x2+9−2*x2)/((x2+9)2)

d(ƒ(x))/d(x)=(9−x2)/((x2+9)2)

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

(9−x2)/((x2+9)2)=0

9−x2=0

x2=9

x=3,x=−3

4. Evaluate the function at the critical points to find the corresponding yvalues.

ƒ(3)=3/(3+9)=3/18=1/6

ƒ*(−3)=(−3)/((−3)2+9)=(−3)/18=−1/6

5. Apply the First Derivative Test to determine the nature of the critical points. For x<−3 ƒ(x)′<0 for −3<x<3 ƒ(x)′>0 for x>3 ƒ(x)′<0

Local Minimum at *x=−3

Local Maximum at *x=3

Final Answer

Local Maxima: *(3,1/6), Local Minima: *(−3,−1/6)

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