Chapter 4.6, Problem 38E

### Calculus: Early Transcendental Fun...

7th Edition
Ron Larson + 1 other
ISBN: 9781337552516

Chapter
Section

### Calculus: Early Transcendental Fun...

7th Edition
Ron Larson + 1 other
ISBN: 9781337552516
Textbook Problem

# Analyzing the Graph of a Trigonometric Function In Exercises 35-42, analyze and sketch a graph of the function over the given interval. Label any intercepts, relative extrema, points of inflection, and asymptotes. Use a graphing utility to verify your results.Function Interval y = 2 ( x − 2 ) + cos x 0 < x < π

To determine

To graph: The function y=2(x2)+cotx

Explanation

Given: The function y=2(xâˆ’2)+cotx over the interval 0â‰¤xâ‰¤Ï€.

Graph: The domain of the function has been provided as [0,Ï€].

Now find the x and y intercepts by equating y and x to zero respectively to obtain:

The x-intercept is (0.28,0) and there are no y-intercepts.

The function has two vertical asymptotes, x=0,x=Ï€ as the cotangent function is not defined at those points.

Now, differentiate the function with respect to x and equate it to zero to obtain the critical points.

2âˆ’csc2x=0cscx=2x=Ï€4,3Ï€4

This gives three test intervals (0,Ï€4),(Ï€4,3Ï€4),(3Ï€4,Ï€).

Let Ï€6âˆˆ(0,Ï€4).

f'(Ï€6)=2âˆ’csc2Ï€6=âˆ’2<0

The function is decreasing on this interval.

Let Ï€2âˆˆ(Ï€4,3Ï€4).

f'(Ï€2)=2âˆ’csc2Ï€2=1>0

The function is increasing on this interval.

Let 7Ï€8âˆˆ(3Ï€4,Ï€).

f'(7Ï€8)=2âˆ’csc27Ï€8=2âˆ’6.828<0

The function is decreasing on this interval.

This implies that the function has a relative minimum at Ï€4 and a relative maximum at 3Ï€4.

Now obtain the value of the function at these critical points.

f(Ï€4)=2((Ï€4)âˆ’2)+cot(Ï€4)=Ï€2âˆ’3f(3Ï€4)=2((3Ï€4)âˆ’2)+cot(3Ï€4)=3Ï€2âˆ’5

Thus the relative minimum is (3Ï€4,3Ï€2âˆ’5) and the relative maximum is (Ï€4,Ï€2âˆ’3)

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