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

Problem

ƒ(x)=sin2(x)+cos(x)

Solution

1. Find the derivative of the function ƒ(x) with respect to x using the chain rule for the first term.

d(ƒ(x))/d(x)=2*sin(x)*cos(x)−sin(x)

2. Set the derivative to zero to find the critical points.

2*sin(x)*cos(x)−sin(x)=0

3. Factor the expression to solve for x

sin(x)*(2*cos(x)−1)=0

4. Solve for x by setting each factor to zero.

sin(x)=0⇒x=n*π

2*cos(x)−1=0⇒cos(x)=1/2⇒x=±π/3+2*n*π

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

d2(ƒ(x))/(d(x)2)=2*cos2(x)−2*sin2(x)−cos(x)

6. Substitute the critical points into the second derivative. For x=2*n*π cos(x)=1 and sin(x)=0

ƒ″*(2*n*π)=2*(1)2−2*(0)2−1=1

Since 1>0 these are local minima. The value is ƒ*(2*n*π)=0+1=1

7. Substitute x=(2*n+1)*π where cos(x)=−1 and sin(x)=0

ƒ″*((2*n+1)*π)=2*(−1)2−2*(0)2−(−1)=3

Since 3>0 these are also local minima. The value is ƒ*((2*n+1)*π)=0−1=−1

8. Substitute x=±π/3+2*n*π where cos(x)=1/2 and sin2(x)=3/4

ƒ″*(±π/3+2*n*π)=2*(1/4)−2*(3/4)−1/2=−1.5

Since −1.5<0 these are local maxima. The value is ƒ*(±π/3+2*n*π)=3/4+1/2=5/4

Final Answer

Local Maxima: 5/4* at *x=±π/3+2*n*π; Local Minima: *1* at *x=2*n*π,−1* at *x=(2*n+1)*π

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