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What are the amplitude, period, and phase shift for the function?

amplitude: 8,000; period: 10; phase shift: 0

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Over this domain, when does the population reach 18,000? 13,000?

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What is the predicted population in 2007? 2010?

In 2007, the predicted population is 4,413. In 2010, the population will be 11,924.

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For the following exercises, suppose a weight is attached to a spring and bobs up and down, exhibiting symmetry.

Suppose the graph of the displacement function is shown in [link] , where the values on the x -axis represent the time in seconds and the y -axis represents the displacement in inches. Give the equation that models the vertical displacement of the weight on the spring.

A graph of a consine function over one period. Graphed on the domain of [0,10]. Range is [-5,5].
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At time = 0, what is the displacement of the weight?

5 in.

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At what time does the displacement from the equilibrium point equal zero?

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What is the time required for the weight to return to its initial height of 5 inches? In other words, what is the period for the displacement function?

10 seconds

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Inverse Trigonometric Functions

For the following exercises, find the exact value without the aid of a calculator.

sin 1 ( cos ( π 6 ) )

π 3

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cos 1 ( tan ( 3 π 4 ) )

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sin ( sec 1 ( 3 5 ) )

No solution

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cot ( sin 1 ( 3 5 ) )

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tan ( cos 1 ( 5 13 ) )

12 5

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sin ( cos 1 ( x x + 1 ) )

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Graph f ( x ) = cos x and f ( x ) = sec x on the interval [ 0 , 2 π ) and explain any observations.

The graphs are not symmetrical with respect to the line y = x . They are symmetrical with respect to the y -axis.

A graph of cosine of x and secant of x. Cosine of x has maximums where secant has minimums and vice versa. Asymptotes at x=-3pi/2, -pi/2, pi/2, and 3pi/2.
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Graph f ( x ) = sin x and f ( x ) = csc x and explain any observations.

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Graph the function f ( x ) = x 1 x 3 3 ! + x 5 5 ! x 7 7 ! on the interval [ 1 , 1 ] and compare the graph to the graph of f ( x ) = sin x on the same interval. Describe any observations.

The graphs appear to be identical.

Two graphs of two identical functions on the interval [-1 to 1]. Both graphs appear sinusoidal.
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Chapter practice test

For the following exercises, sketch the graph of each function for two full periods. Determine the amplitude, the period, and the equation for the midline.

f ( x ) = 0.5 sin x

amplitude: 0.5; period: 2 π ; midline y = 0

A graph of two periods of a sinusoidal function, graphed over -2pi to 2pi. The range is [-0.5,0.5]. X-intercepts at multiples of pi.
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f ( x ) = 5 sin x

amplitude: 5; period: 2 π ; midline: y = 0

Two periods of a sine function, graphed over -2pi to 2pi. The range is [-5,5], amplitude of 5, period of 2pi.
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f ( x ) = cos ( x + π 3 ) + 1

amplitude: 1; period: 2 π ; midline: y = 1

A graph of two periods of a cosine function, graphed over -7pi/3 to 5pi/3. Range is [0,2], Period is 2pi, amplitude is1.
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f ( x ) = 5 sin ( 3 ( x π 6 ) ) + 4

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f ( x ) = 3 cos ( 1 3 x 5 π 6 )

amplitude: 3; period: 6 π ; midline: y = 0

A graph of two periods of a cosine function, over -7pi/2 to 17pi/2. The range is [-3,3], period is 6pi, and amplitude is 3.
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f ( x ) = 2 tan ( x 7 π 6 ) + 2

amplitude: none; period:   π ;   midline:   y = 0 , asymptotes:   x = 2 π 3 + π k , where   k   is an integer

A graph of two periods of a tangent function over -5pi/6 to 7pi/6. Period is pi, midline at y=0.
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f ( x ) = π cos ( 3 x + π )

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f ( x ) = 5 csc ( 3 x )

amplitude: none; period:   2 π 3 ;   midline:   y = 0 , asymptotes:   x = π 3 k , where   k   is an integer

A graph of two periods of a cosecant functinon, over -2pi/3 to 2pi/3. Vertical asymptotes at multiples of pi/3. Period of 2pi/3.
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f ( x ) = π sec ( π 2 x )

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f ( x ) = 2 csc ( x + π 4 ) 3

amplitude: none; period: 2 π ; midline: y = 3

A graph of two periods of a cosecant function, graphed from -9pi/4 to 7pi/4. Period is 2pi, midline at y=-3.
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For the following exercises, determine the amplitude, period, and midline of the graph, and then find a formula for the function.

Give in terms of a sine function.

A graph of two periods of a sine function, graphed from -2 to 2. Range is [-6,-2], period is 2, and amplitude is 2.
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Give in terms of a sine function.

A graph of two periods of a sine function, graphed over -2 to 2. Range is [-2,2], period is 2, and amplitude is 2.

amplitude: 2; period: 2; midline: y = 0 ; f ( x ) = 2 sin ( π ( x 1 ) )

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Give in terms of a tangent function.

A graph of two periods of a tangent function, graphed over -3pi/4 to 5pi/4. Vertical asymptotes at x=-pi/4, 3pi/4. Period is pi.
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For the following exercises, find the amplitude, period, phase shift, and midline.

y = sin ( π 6 x + π ) 3

amplitude: 1; period: 12; phase shift: −6 ; midline y = −3

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y = 8 sin ( 7 π 6 x + 7 π 2 ) + 6

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The outside temperature over the course of a day can be modeled as a sinusoidal function. Suppose you know the temperature is 68°F at midnight and the high and low temperatures during the day are 80°F and 56°F, respectively. Assuming t is the number of hours since midnight, find a function for the temperature, D , in terms of t .

D ( t ) = 68 12 sin ( π 12 x )

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Water is pumped into a storage bin and empties according to a periodic rate. The depth of the water is 3 feet at its lowest at 2:00 a.m. and 71 feet at its highest, which occurs every 5 hours. Write a cosine function that models the depth of the water as a function of time, and then graph the function for one period.

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For the following exercises, find the period and horizontal shift of each function.

g ( x ) = 3 tan ( 6 x + 42 )

period: π 6 ; horizontal shift: −7

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n ( x ) = 4 csc ( 5 π 3 x 20 π 3 )

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Write the equation for the graph in [link] in terms of the secant function and give the period and phase shift.

A graph of 2 periods of a secant function, graphed over -2 to 2. The period is 2 and there is no phase shift.

f ( x ) = sec ( π x ) ; period: 2; phase shift: 0

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If tan x = 3 , find tan ( x ) .

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If sec x = 4 , find sec ( x ) .

4

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For the following exercises, graph the functions on the specified window and answer the questions.

Graph m ( x ) = sin ( 2 x ) + cos ( 3 x ) on the viewing window [ 10 , 10 ] by [ 3 , 3 ] . Approximate the graph’s period.

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Graph n ( x ) = 0.02 sin ( 50 π x ) on the following domains in x : [ 0 , 1 ] and [ 0 , 3 ] . Suppose this function models sound waves. Why would these views look so different?

The views are different because the period of the wave is 1 25 . Over a bigger domain, there will be more cycles of the graph.

Two side-by-side graphs of a sinusodial function. The first graph is graphed over 0 to 1, the second graph is graphed over 0 to 3. There are many periods for each.
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Graph f ( x ) = sin x x on [ 0.5 , 0.5 ] and explain any observations.

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For the following exercises, let f ( x ) = 3 5 cos ( 6 x ) .

What is the largest possible value for f ( x ) ?

3 5

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What is the smallest possible value for f ( x ) ?

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Where is the function increasing on the interval [ 0 , 2 π ] ?

On the approximate intervals ( 0.5 , 1 ) , ( 1.6 , 2.1 ) , ( 2.6 , 3.1 ) , ( 3.7 , 4.2 ) , ( 4.7 , 5.2 ) , ( 5.6 , 6.28 )

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For the following exercises, find and graph one period of the periodic function with the given amplitude, period, and phase shift.

Sine curve with amplitude 3, period π 3 , and phase shift ( h , k ) = ( π 4 , 2 )

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Cosine curve with amplitude 2, period π 6 , and phase shift ( h , k ) = ( π 4 , 3 )

f ( x ) = 2 cos ( 12 ( x + π 4 ) ) + 3

A graph of one period of a cosine function, graphed over -pi/4 to 0. Range is [1,5], period is pi/6.
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For the following exercises, graph the function. Describe the graph and, wherever applicable, any periodic behavior, amplitude, asymptotes, or undefined points.

f ( x ) = 5 cos ( 3 x ) + 4 sin ( 2 x )

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f ( x ) = e sin t

This graph is periodic with a period of 2 π .

A graph of two periods of a sinusoidal function, The graph has a period of 2pi.
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For the following exercises, find the exact value.

cos 1 ( sin ( π ) )

π 2

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cos 1 ( tan ( 7 π 4 ) )

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cos ( sin 1 ( 1 2 x ) )

1 ( 1 2 x ) 2

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cos ( tan 1 ( x 2 ) )

1 1 + x 4

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For the following exercises, suppose sin t = x x + 1 .

Given [link] , find the measure of angle θ to three decimal places. Answer in radians.

An illustration of a right triangle with angle theta. Opposite the angle theta is a side with length 12, adjacent to the angle theta is a side with length 19.
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For the following exercises, determine whether the equation is true or false.

arcsin ( sin ( 5 π 6 ) ) = 5 π 6

False

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arccos ( cos ( 5 π 6 ) ) = 5 π 6

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The grade of a road is 7%. This means that for every horizontal distance of 100 feet on the road, the vertical rise is 7 feet. Find the angle the road makes with the horizontal in radians.

approximately 0.07 radians

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Questions & Answers

for the "hiking" mix, there are 1,000 pieces in the mix, containing 390.8 g of fat, and 165 g of protein. if there is the same amount of almonds as cashews, how many of each item is in the trail mix?
ADNAN Reply
linear speed of an object
Melissa Reply
an object is traveling around a circle with a radius of 13 meters .if in 20 seconds a central angle of 1/7 Radian is swept out what are the linear and angular speed of the object
Melissa
test
Matrix
how to find domain
Mohamed Reply
like this: (2)/(2-x) the aim is to see what will not be compatible with this rational expression. If x= 0 then the fraction is undefined since we cannot divide by zero. Therefore, the domain consist of all real numbers except 2.
Dan
define the term of domain
Moha
if a>0 then the graph is concave
Angel Reply
if a<0 then the graph is concave blank
Angel
what's a domain
Kamogelo Reply
The set of all values you can use as input into a function su h that the output each time will be defined, meaningful and real.
Spiro
how fast can i understand functions without much difficulty
Joe Reply
what is inequalities
Nathaniel
functions can be understood without a lot of difficulty. Observe the following: f(2) 2x - x 2(2)-2= 2 now observe this: (2,f(2)) ( 2, -2) 2(-x)+2 = -2 -4+2=-2
Dan
what is set?
Kelvin Reply
a colony of bacteria is growing exponentially doubling in size every 100 minutes. how much minutes will it take for the colony of bacteria to triple in size
Divya Reply
I got 300 minutes. is it right?
Patience
no. should be about 150 minutes.
Jason
It should be 158.5 minutes.
Mr
ok, thanks
Patience
100•3=300 300=50•2^x 6=2^x x=log_2(6) =2.5849625 so, 300=50•2^2.5849625 and, so, the # of bacteria will double every (100•2.5849625) = 258.49625 minutes
Thomas
158.5 This number can be developed by using algebra and logarithms. Begin by moving log(2) to the right hand side of the equation like this: t/100 log(2)= log(3) step 1: divide each side by log(2) t/100=1.58496250072 step 2: multiply each side by 100 to isolate t. t=158.49
Dan
what is the importance knowing the graph of circular functions?
Arabella Reply
can get some help basic precalculus
ismail Reply
What do you need help with?
Andrew
how to convert general to standard form with not perfect trinomial
Camalia Reply
can get some help inverse function
ismail
Rectangle coordinate
Asma Reply
how to find for x
Jhon Reply
it depends on the equation
Robert
yeah, it does. why do we attempt to gain all of them one side or the other?
Melissa
how to find x: 12x = 144 notice how 12 is being multiplied by x. Therefore division is needed to isolate x and whatever we do to one side of the equation we must do to the other. That develops this: x= 144/12 divide 144 by 12 to get x. addition: 12+x= 14 subtract 12 by each side. x =2
Dan
whats a domain
mike Reply
The domain of a function is the set of all input on which the function is defined. For example all real numbers are the Domain of any Polynomial function.
Spiro
Spiro; thanks for putting it out there like that, 😁
Melissa
foci (–7,–17) and (–7,17), the absolute value of the differenceof the distances of any point from the foci is 24.
Churlene Reply
Practice Key Terms 6

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Source:  OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6
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