Graph one period of f⁡(x)=0.5⁢csc⁡(2⁢x)

. We can use two reference points, the local maximum at (π8,−3)

Graph one period of f⁡(x)=−6⁢sec⁡(4⁢x+2)−8

. There is a local minimum at (1.5,5)

Graph one period of y=4⁢sec⁡(π3⁢x−π2)+1

This is a vertical reflection of the preceding graph because A

Find a formula for the function in Figure 6.2.7

Plot reference points at (P4,A), (0,0), and (−P4,−A), and draw the graph through these points.

ecall that, for a point on a circle of radius r, the y-coordinate of the point i

. The phase shift is −2

um at x=1 and the maximum

quarter points inc

possibilities

function

shift

units

, which shifts to the right by π4

If C<0, the graph shifts to the left.

xis, bec

find

functions

If cos(t)=1213 cos(t)=1213 and t t is in quadrant IV, as shown in Figure 5.3.8

Now let’s take a moment to reconsider the Ferris wheel introduced at the beginning of this section. Suppose a rider snaps a photograph while stopped twenty feet above ground level. The rider then rotates three-quarters of the way around the circle. What is the rider’s new elevation? To answer questions such as this one, we need to evaluate the sine or cosine functions at angles that are greater than 90 degrees or at a negative angle. Reference angles make it possible to evaluate trigonometric functions for angles outside the first quadrant. They can also be used to find (x,y)

Looking for a thrill? Then consider a ride on the Singapore Flyer, the world’s tallest Ferris wheel. Located in Singapore, the Ferris wheel soars to a height of 541 feet—a little more than a tenth of a mile! Described as an observation wheel, riders enjoy spectacular views as they travel from the ground to the peak and down again in a repeating pattern. In this section, we will examine this type of revolving motion around a circle. To do so, we need to define the type of circle first, and then place that circle on a coordinate system. Then we can discuss circular motion in terms of the coordinate pairs.

Next, we will find the cosine and sine of the reference angle: cos⁡(π6)=32        sin⁡(π6)=12

a

is the same length, and we know one side is the radius of the unit circle, all sides must be of length 1.

Divide the total rotation in radians by the elapsed time to find the angular speed: apply ω=θt

A satellite is rotating around Earth at 0.25 radians per hour at an altitude of 242 km above Earth. If the radius of Earth is 6378 kilometers, find the linear speed of the satellite in kilometers per hour.

Angular speed can be given in radians per second, rotations per minute, or degrees per hour for example.

Notice what happens if we find the ratio of the arc length divided by the radius of the circle. (5.1.10)Smaller circle: 12⁢π2=14⁢π(5.1.11)Larger circle: 34⁢π3=14⁢π

As with exponential models, data modeled by logarithmic functions are either always increasing or always decreasing as time moves forward.

4.7: Exponential and Logarithmic Models

Rewrite y=a⁢bx as y=a⁢eln⁡(bx). Use the power rule of logarithms to rewrite y as y=a⁢ex⁢ln⁡(b)=a⁢eln⁡(b)x. Note that a=A0 and k=ln⁡(b) in the equation y=A0⁢ek⁢x.

Does a linear, exponential, or logarithmic model best fit the data in Table 4.7.2

This formula is derived as follows: T⁡(t)=A⁢bc⁢t+TsT⁡(t)=A⁢eln⁡(bc⁢t)+TsLaws of logarithmsT⁡(t)=A⁢ec⁢t⁢ln⁡b+TsLaws of logarithmsT⁡(t)=A⁢ek⁢t+TsRename the constant c l⁢n⁢b, calling it k

ht of 12 an

The formula is derived as follows 12⁢A0=A0⁢ek⁢t12=ek⁢tDivide by A0ln⁡(12)=k⁢t⁢vTake the natural log−ln⁡(2)=k⁢tApply laws of logarithms−ln⁡(2)⁢k=tDivide by k

he labeled points (1/k, (A_0)e), (0, A_0), and (-1/k, (A_0)/e). The second graph is of when k<0 and with the labeled points (-1/k, (A_0)e), (0, A_0), and (1/k, (A_0)/e)." src="/@api/deki/files/12238/fig_6.8.4.jpg">

as we can see in Figure 4.7.2 and Figure 4.7.3.

We can see how widely the half-lives for these substances vary. Knowing the half-life of a substance allows us to calculate the amount remaining after a specified time. We can use the formula for radioactive decay: (4.6.4)A⁡(t)=A0⁡eln⁡(0.5)T⁢t(4.6.5)A⁡(t)=A0⁡eln⁡(0.5)⁢tT(4.6.6)A⁡(t)=A0⁡(eln⁡(0.5))tT(4.6.7)A⁡(t)=A0⁡(12)tT

takes for half of the u

Solve

Equations

Key Equations

For example, to evaluate log⁡(100), we can rewrite the logarithm as log10⁡(102), and then apply the inverse property logb⁡(bx)=x to get log10⁡(102)=2. To evaluate eln⁡(7), we can rewrite the logarithm as eloge⁡7, and then apply the inverse property blogb⁡x=x to get eloge⁡7=7. Finally, we have the one-to-one property.

Change

Condense

Expand

rough the points (–1,1)

and has been vertically reflected. We do not know yet the vertical shift or the vertical stretch. We know so far that the equation will have form:

4.4: Graphs of Logarithmic Functions

and key point (b,1)

notice

Label the three points.

Consider the three key points from the parent function, (13,−1)

Identify three key points from the parent function. Find new coordinates for the shifted functions by adding d

Consider the three key points from the parent function, (13,−1)

4.3: Logarithmic Functions

is equivalent to log2⁡(12)=−1

wo other points

Plot the y-intercept, (0,1),along with two other points. We can use (−1,4) and (1,0.25). Dra

Plot the key point (b,1)

days

square root is only defined when the quantity under the radical is non-negative,

y⁡(x)=12⁢x2 We are interested in the surface area of the water, so we must determine the width at the top of the water as a function of the water depth. For any depth y, the width will be given by 2⁢x, so we need to solve the equation above for x and find the inverse function. However, notice that the original function is not one-to-one, and indeed, given any output there are two inputs that produce the same output, one positive and one negative. To find an inverse, we can restrict our original function to a limited domain on which it is one-to-one. In this case, it makes sense to restrict ourselves to positive x values. On this domain, we can find an inverse by solving for the input variable:

is exhibiting a behavior similar to 1x2

The one at x=–1 seems to exhibit the basic behavior similar to 1x, with the graph heading toward positive infinity on one side and heading toward negative infinity on the other. The asymptote at x=2 is exhibiting a behavior similar to 1x2, with the graph heading toward negative infinity on both sides of the asymptote. See Figure 3.7.25.

Writing Rational Functions

o we know the behavior will be the same on both sides of the asymptote.

or the vertical asymptote at x=2, the factor was not squared, so the graph will have opposite behavior on either side of the asymptote.

The factor associated with the vertical asymptote at x=−1 was squared, so we know the behavior will be the same on both sides of the asymptote. The graph heads toward positive infinity as the inputs approach the asymptote on the right, so the graph will head toward positive infinity on the left as well.

When the degree of the factor in the denominator is even, the distinguishing characteristic is that the graph either heads toward positive infinity on both sides of the vertical asymptote or heads toward negative infinity on both sides. See Figure 3.7.19.

e degree of the factor in the denominator is odd, the distinguishing characteristic

Figure

This function will have a horizontal asymptote at y=0. See Figure 3.7.16.

Likewise, a rational function’s end behavior will mirror that of the ratio of the function that is the ratio of the leading terms.

The vertical asymptotes of a rational function may be found by examining the factors of the denominator that are not common to the factors in the numerator. Vertical asymptotes occur at the zeros of such factors.

To find the horizontal asymptote, divide the leading coefficient in the numerator by the leading coefficient in the denominator: (3.7.9)110=0.1

As the inputs increase and decrease without bound, the graph appears to be leveling off at output values of 3, indicating a horizontal asymptote at y=3.

351=13⁢w3+43⁢w2

We can conclude if k is a zero of f⁡(x), then x−k is a factor of f⁡(x).

We can apply this theorem to a special case that is useful in graphing polynomial functions. If a point on the graph of a continuous function f at x=a lies above the x-axis and another point at x=b lies below thex-axis, there must exist a third point between x=a and x=b where the graph crosses the x-axis. Call this point (c,f⁡(c)).This means that we are assured there is a solution c where f⁡(c)=0.

distance

Figure

intercept

coordinates

Now that we can find the inverse of a function, we will explore the graphs of functions and their inverses. Let us return to the quadratic function f⁡(x)=x2

Solution

Identify which of the toolkit functions besides the quadratic function are not one-to-one, and find a restricted domain on which each function is one-to-one, if any. The toolkit functions are reviewed in Table 1.7.2

the identity function

Table 1.6.1

Find the domain of (f∘g)⁡(x) where f⁡(x)=1x−2 and g⁡(x)=x+4

Suppose f⁡(x) gives miles that can be driven in x hours and g⁡(y) gives the gallons of gas used in driving y miles. Which of these expressions is meaningful: f⁡(g⁡(y)) or g⁡(f⁡(x))? Solution The function y=f⁡(x) is a function whose output is the number of miles driven corresponding to the number of hours driven. number of miles =f⁡(number of hours) The function g⁡(y) is a function whose output is the number of gallons used corresponding to the number of miles driven. This means: number of gallons =g⁡(number of miles) The expression g⁡(y) takes miles as the input and a number of gallons as the output. The function f⁡(x) requires a number of hours as the input. Trying to input a number of gallons does not make sense. The expression f⁡(g⁡(y)) is meaningless. The expression f⁡(x) takes hours as input and a number of miles driven as the output. The function g⁡(y) requires a number of miles as the input. Using f⁡(x) (miles driven) as an input value for g⁡(y), where gallons of gas depends on miles driven, does make sense. The expression g⁡(f⁡(x)) makes sense, and will yield the number of gallons of gas used, g, driving a certain number of miles, f⁡(x), in x hours.

a+7

people, n, to the cost, C.

The vertical extent of the graph is all range values 5 and below,

If the function’s formula contains an even root, set the radicand greater than or equal to 0, and then solve.

solve for x

interval that is more than 0 and less than or equal to 100 and write (0,100].

Domain and Range

for any input, r, there is only one output, A.

inputs q and r both give output n.

Figures 1.1.1⁢a and 1.1.1⁢b.

To solve f⁡(x)=4, we find the output value 4 on the vertical axis. Moving horizontally along the line y=4, we locate two points of the curve with output value 4: (−1,4) and (3,4). These points represent the two solutions to f⁡(x)=4: −1 or 3. This means f⁡(−1)=4 and f⁡(3)=4, or when the input is −1 or 3, the output is 4. See Figure 1.1.9.

How To: Given a function represented by a table, identify specific output and input values 1. Find the given input in the row (or column) of input values. 2. Identify the corresponding output value paired with that input value. 3. Find the given output values in the row (or column) of output values, noting every time that output value appears. 4. Identify the input value(s) corresponding to the given output value.

There is an urban legend that a goldfish has a memory of 3 seconds, but this is just a myth. Goldfish can remember up to 3 months, while the beta fish has a memory of up to 5 months. And while a puppy’s memory span is no longer than 30 seconds, the adult dog can remember for 5 minutes. This is meager compared to a cat, whose memory span lasts for 16 hours.

Puppy 0.008 Adult Dog 0.083 Cat 3 Goldfish 2160 Beta Fish

We can rewrite it to decide if p is a function of n.

and h⁡(4)=24

With an input value of a+h, we must use the distributive property.

b. In this case, the input value is a letter so we cannot simplify the answer any further. f⁡(a)=a2+3⁢a−4

Table 1.1.5 displays the age of children in years and their corresponding heights.