Find the Local Maxima and Minima f(x)=x+2cos(x)

Problem

ƒ(x)=x+2*cos(x)

Solution

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

d(ƒ(x))/d(x)=1−2*sin(x)

2. Set the derivative to zero to solve for the critical values of x

1−2*sin(x)=0

sin(x)=1/2

3. Determine the general solutions for x based on the unit circle.

x=π/6+2*n*π

x=(5*π)/6+2*n*π

4. Find the second derivative to apply the Second Derivative Test for concavity.

d2(ƒ(x))/(d(x)2)=−2*cos(x)

5. Test the first set of points x=π/6+2*n*π in the second derivative.

ƒ″*(π/6+2*n*π)=−2*cos(π/6)=−2*(√(,3)/2)=−√(,3)

Since ƒ(x)″<0 these points are local maxima.

6. Test the second set of points x=(5*π)/6+2*n*π in the second derivative.

ƒ″*((5*π)/6+2*n*π)=−2*cos((5*π)/6)=−2*(−√(,3)/2)=√(,3)

Since ƒ(x)″>0 these points are local minima.

7. Calculate the y-coordinates for the local extrema.

ƒ*(π/6+2*n*π)=π/6+2*n*π+√(,3)

ƒ*((5*π)/6+2*n*π)=(5*π)/6+2*n*π−√(,3)

Final Answer

Local Maxima: *(π/6+2*n*π,π/6+2*n*π+√(,3)), Local Minima: *((5*π)/6+2*n*π,(5*π)/6+2*n*π−√(,3))

Want more problems? Check here!