# The local maximum and minimum and the absolute maximum and absolute minimum of the function f ( x ) = x + 2 cos x on the interval [ − π , π ] .

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter 4, Problem 5RE
To determine

## To find: The local maximum and minimum and the absolute maximum and absolute minimum of the function f(x)=x+2cosx on the interval [−π,π].

Expert Solution

The local maximum occurs at x=π6.

The local minimum occurs at x=5π6.

The absolute maxima occurs at x=π6.

The absolute minimum is x=π.

### Explanation of Solution

Given:

The function is, f(x)=x+2cosx and the interval is, [π,π].

Calculation:

Obtain the first derivative of the given function.

f'(x)=12sinx

Set f(x)=0 and obtain the critical numbers.

12sinx=01=2sinxsinx=12sin(π6)=12

Also, it is knows that sin(πθ)=θ and hence sin(ππ6)=π6.

Therefore, x=5π6.

Thus, the critical numbers are x=π6 and x=5π6 which lies on the given interval [π,π].

Apply the extreme values of the given interval and the critical numbers in f(x).

Substitute x=π in f(x),

f(π)=π+2cos(π)=π23.1425.14

Substitute x=π6 in f(x),

f(π6)=π6+2cos(π6)=π6+232=3.146+32.256

Substitute x=5π6 in f(x),

f(5π6)=5π6+2cos(5π6)=5π6+2(32)=5(3.14)630.886

Substitute x=π in f(x),

f(π)=π+2cos(π)=π23.1421.14

Since the largest functional value is the absolute maximum and the smallest functional value is the absolute minimum, the absolute maximum of f(x) is approximately 2.256 and the absolute minimum of f(x) is −5.14.

Therefore, the absolute maxima of f(x) occurs at x=π6 and the absolute minimum of f(x) is occurs at x=π.

To find the local maximum and minimum obtain the second derivative of the function.

f''(x)=ddx(12sinx)=2cosx

Substitute the critical numbers in the second derivative and obtain the local maximum and minimum as follows.

Substitute x=π6 in f''(x),

f''(π6)=2cos(π6)=2(32)=3<0

Substitute x=5π6 in f''(x),

f''(5π6)=2cos(5π6)=2(32)=3>0

Thus, the local maximum occurs at x=π6 and the local minimum occurs at x=5π6.

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