Find the Local Maxima and Minima f(x)=9sin(x)+9cos(x)

Problem

ƒ(x)=9*sin(x)+9*cos(x)

Solution

1. Find the first derivative of the function to identify critical points.

d(ƒ(x))/d(x)=9*cos(x)−9*sin(x)

2. Set the derivative to zero to solve for x within the standard period [0,2*π)

9*cos(x)−9*sin(x)=0

cos(x)=sin(x)

tan(x)=1

x=π/4,(5*π)/4

3. Find the second derivative to apply the Second Derivative Test.

d^2(ƒ(x))/(d(x)^2)=−9*sin(x)−9*cos(x)

4. Evaluate the second derivative at the critical points to determine concavity.

d^2(ƒ(π/4))/(d(x)^2)=−9*sin(π/4)−9*cos(π/4)=−9√(,2)<0

d^2(ƒ((5*π)/4))/(d(x)^2)=−9*sin((5*π)/4)−9*cos((5*π)/4)=9√(,2)>0

5. Calculate the function values at these points. Since the second derivative is negative at x=π/4 it is a local maximum. Since it is positive at x=(5*π)/4 it is a local minimum.

ƒ(π/4)=9*(√(,2)/2)+9*(√(,2)/2)=9√(,2)

ƒ((5*π)/4)=9*(−√(,2)/2)+9*(−√(,2)/2)=−9√(,2)

Final Answer

Local Maxima: *(2*n*π+π/4,9√(,2)), Local Minima: *(2*n*π+(5*π)/4,−9√(,2))

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