   Chapter 3.3, Problem 48E

Chapter
Section
Textbook Problem

# 47-48(a) Use a graph of f to estimate the maximum and minimum values. Then find the exact values.(b) Estimate the value of x at which f increases most rapidly. Then find the exact value. f ( x ) = x + 2 cos x ,   0 ≤ x ≤ 2 π

To determine

(a)

To estimate:

The maximum and minimum values by using graph of f and find exact values.

Explanation

1) Concept:

(i) To find local minimum and local maximum from graph find highest and lowest point on graph if it exists.

(ii) Extremum values of functions are obtained by using f'c=0,

Increasing or Decreasing test:

If f'x>0 then f is increasing

If f'x<0 then f is decreasing

2) Given:

fx=x+2cosx

3) Calculation:

(i) Consider fx=x+2cosx,

Therefore, graph of given function is

From the graph,

On interval [0, 2π] minimum value of function is at x=2.61.

Also graph changes from decreasing to increasing, Local and absolute minimum both are same which is f2.610.89,

Similarly on interval [0, 2π] maximum value of function is at x=2π. Hence, absolute maximum is f2π8.28 and graph of function changes from increasing to decreasing there is local maxima at x=0.53, local maximum value f0.53=2.26.

(ii) For finding exact values,

Consider, fx=x+2cosx

Differentiating fx with respect to x,

f'x=1-2sinx

For finding critical points,f'x=0,

Therefore, 1-2sinx=0

sinx=12

x=π6 and x=5π6

Now domain of above function is [0, 2π]

We split domain into the intervals according to critical numbers

By using Increasing or decreasing test,

On interval 0,π6,f'π12=1-2sinπ12=0.4424>0 Therefore, f is increasing

On interval π6,5π6,f'π2=1-2sinπ2=1-2<0 Therefore, f is decreasing

On interval 5π6,2π, f'π=1-2sinπ=1>0 therefore, f is increasing

As f is increasing on interval 0,π6 and decreasing on the interval π6,5π6

To determine

(b)

To estimate:

The value of x at which f increases more rapidly and find the exact values

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