Section 4-4:Periodic Functions - Stretching and Translating
A function f is periodic if there is a positive number p
such that:
f(x + p) = f(x)
for all x in the domain of f.
The definition means that the y values will repeat over some p value called the fundamental period of the
function. Look at the graph:
The graph starts at 0, goes up to 2, back down to 0, down to -2 and back to 0. At this point the graph starts repeating. Look at the x value to find the period length. The period length p = 4, because it takes 4 units for this graph to repeat the y values. You can tell where the graph will be at larger values by knowing the repeat. f(0) = 0, f(1) = 2, f(2) = 0, f(3) = -2 and f(4) =
0. What is f(21)? Divide 21 by 4 and use the remainder. The remainder is 1. Thus f(21) = f(1) = 2
What is f(82)? Divide 82 by 4. The remainder is
2. Thus f(82) = f(2) = 0.
If a periodic graph has a maximum value M and a minimum value m, then the amplitude A of the function
is:
A = (M - m)/2
The amplitude of the above graph is:
A = (2 -(-2))/2 = 4/2 = 2
Stretching a graph vertically
The graph of y = cf(x) where c is a positive number not equal to 1, is obtained by vertically stretching or
shrinking the graph of y = f(x). Let f(x) be the following graph:
Now compare these two graphs to the green one above.
The purple graph doubled the green graph. Notice, all
that changed was the high and low points of the graph. In other words, stretched vertically. Look at the red graph. It is the green graph multiplied by 1/2. All that has changed is the high and low points. In other words, shrunk vertically. The period length has not changed. In all three graphs, the period
length is 3.
Horizontally stretching or shrinking
The graph of y = f(cx) where c is positive and not equal
to one is obtained by horizontally stretching or shrinking the graph of y = f(x). If c > 1 it is a horizontal shrink. If 0 < c < 1, it is a horizontal
stretch.
Here is y = f(x):
Watch the effect of multiplying the x value by 2.
Notice the amplitude didn't change. The graph high and low points are the same. But look at the purple graph. Notice it has gone through two complete cycles by the time the green graph has gone through one
cycle. It is like compressing a spring. Now watch the effect of multiplying by 1/2.
The effect this time is to stretch the graph. Look at the red graph. At x = 3, the red graph is only half way through the cycle. It takes 6 units for the red graph to repeat instead of three for the green graph. It is like
pulling out on a spring.
Summary of above
If a periodic function f has a period p and amplitude
A, then:
y = cf(x) has period p and amplitude cA
y = f(cx) has period p/c and amplitude A
Translating Graphs
A translation is simply moving the exact same graph to another location. The size and shape does not change from the original graph, only the placement of the graph changes. Your knowledge of basic graphs is very helpful when doing translations. Here is how to
translate:
y - k = f(x - h) is obtained by shifting the graph of y =
f(x), k units up/down and h units right/left.
You already know what y = x2 is. How does y - 2 = (x -
1)2 compare?
Notice the green graph is the same size and shape of the blue graph. It is shifted one unit right and two units
up.
Now graph y + 1 = | x + 2|. This depends on you knowing that the absolute value graph is a v-shaped
graph. So this is a translation of y = |x|
The green graph is the graph of the function we want. It is a translation of the blue graph moved one unit down and 2 units left. Notice in this problem and the last problem what causes the graph to be shifted
right vs. left and up vs. down.
Combining reflections and translations
When combining reflections and translations, remember to reflect first then translate. Failure to work the problem in this order may result in the wrong
answer.
Graph the function y - 1 = -|x + 1|
The basic graph is y = |x|, a v-shaped graph. The negative sign in front makes this a reflection about the x-axis. Do this first. Then translate the result by moving the graph up one and one to the left. Our
answer is in purple.
Remember to make a copy of the chart on page 142 in your notebook. You must know that chart!! It helps a
great deal in future chapters!!
Let's shift into the next section:
Let's reflect back on the previous section:
Section 4-5: Inverse Functions
An inverse is the operation that takes you back to where you started. The inverse of multiplication is division, adding and subtracting, square and square root, etc. For functions, there are two conditions for a function to
be the inverse function:
1) g(f(x)) = x for all x in the domain of f
2) f(g(x)) = x for all x in the domain of g
Notice in both cases you will get back to the element
that you started with, namely, x.
The notation used to indicate an inverse function is: f
-1
(x) pronounced \"f inverse\". This notation does not
mean 1/f(x).
Example
1) If f(x) = 3x - 1 and g(x) = (x + 1)/3, show that f and
g are inverses to each other.
To show that they are inverses, we must prove
both of the above parts.
g(f(x)) = g(3x - 1) = (3x - 1 + 1)/3 = 3x/3 =
x
f(g(x)) = f((x + 1)/3) = 3[(x + 1)/3] - 1 = x
+ 1 - 1 = x
Since both parts work, they are indeed inverses of each
other.
To find a rule for the inverse function
Find the inverse function for y = 5x + 2
To find the inverse, interchange x and y.
x = 5y + 2
Now isolate for y!!
x - 2 = 5y (x - 2)/5 = y
We now have the inverse!!
Notice, that this inverse make sense. The original problem had adding by two and the inverse is subtracting two. The original function had multiplying
by five and the inverse has division by five.
Graphs of inverse functions
We have to make sure that the inverse is indeed a function. Not all functions will have inverses that are also functions. In order for a function to have an inverse, it must pass the horizontal line test!!
Horizontal line test
If the graph of a function y = f(x) is such that no horizontal line intersects the graph in more than one
point, then f has an inverse function.
This will make sense when we discover how to graph the inverse function. To graph the inverse function, it is simply the reflection about the line y = x. Makes sense, because in order to get the graph, we interchange
x and y. Recall from previous sections what the reflection about the line y = x looks like. Any two points on the same horizontal line when reflected will be on the same vertical line. Can't have this because it wouldn't be a function. That's why the horizontal line
test works.
Example
1) Find the equation of f -1 and graph f, f -1, and y = x
for f(x) = 2x - 5.
First, f(x) is a line and it passes the horizontal line test.
Find the inverse: y = 2x - 5 x = 2y - 5 x + 5 = 2y (x + 5)/2 = y
2) Let f(x) = 9 - x2 for x > 0
Find the equation for f -1(x)
Sketch the graph of f, f -1, and y = x.
Notice that the equation is half of a parabola. Only the side to the right of zero. If we tried to use the entire parabola, it wouldn't pass the horizontal line test.
To find the equation: y = 9 - x2 x = 9 - y2 x - 9 = -y2
9 - x = y2
, x < 9
On to the last section: Back up and regroup:
Section 4-6: Functions of Two Variables
Try the quiz at the bottom of the page!
go to quiz
Suppose a company makes two products. Let x represent the number of the first product made and y represent the number of the second product made. If the first product makes $2 each and the second $3 each,
the profit function would be:
P(x, y) = 2x + 3y
Notice that profit is a function of two variables. There are a few ways we could graph this function. One way is to graph in a three dimensional plane. We will talk about this later in the year. Another way to graph this function is to hold one of the variables constant. For example, we could hold y constant like: y = 2, or y = 3
or y = 4, etc. The equations would look like:
P(x, 2) = 2x + 6 P(x, 3) = 2x + 9 P(x, 4) = 2x + 12
These graphs all have the same slope but different y-intercepts. These make a family of curves. They will
be graphed in the xP plane.
Another way we can graph this is by holding the profit constant. For example, we could let P = 6 or P = 12 or
P = 18. Then the equations would be:
6 = 2x + 3y 12 = 2x + 3y 18 = 2x + 3y
Another equation we could use is:
d(r, t) = rt
Suppose we hold t constant at t = 1, t = 2 and t =
3. Make a graph of these three equations:
d(r, 1) = r d(r, 2) = 2r d(r, 3) = 3r
Now graph the same function holding the distance
constant at d = 10, d = 20, and d = 30. The equations are
10 = rt 20 = rt 30 = rt
I'm ready for that practice test now!
I'm almost ready. I think I will go back and
review!
Current quizaroo # 4b
1) y + 3 = -f(x - 1) Describe the translations, symmetry and stretching/shrinking for this graph:
a) Reflected about the y-axis, moved three units up and one unit to the left
b) Reflected about the y-axis, moved three units down and one unit to the right
c) Reflected about the x-axis, moved three units up and one unit to the left
d) Reflected about the x-axis, moved three units down and one unit to the right
e) Moved three units down and one unit to the right
2) Write the equation of a function that is shifted 2 units up and 4 units to the left, that is reflected about the x-axis with a stretch vertical factor of 2.
a) y - 2 = -2f(x + 4)
b) y + 2 = -2f(x - 4) c) y - 2 = 2f(x + 4)
d) y + 2 = 2f(x - 4) e) y +4 = -2f(x - 2)
3) Find the inverse function to the equation: y = -3x + 4
a) y = 3x - 4 b) y = (4 - x)/3 c) y = (x - 4)/3 d) y = (x + 3)/4 e) y = (3 - x)/4
4) If g(x) = x2 - 2, x < 0, find the inverse function. ____
a) y = -\\/ x + 2, x > -2 ____
b) y = \\/ x + 2, x > -2
c) y = x + 2, x > -2
d) y = -(x + 2), x > -2
e) doesn't have an inverse function, because it doesn't pass the
horizontal line test
5) Name an ordered pair that is not in the domain of f(a, b) = (a - b)/(a + b)
a) b) c) d) e)
(5, 5) (x, x) (-d, d) (-4, -4) (3/2, 2/3)
click here for answers!!
Sample test!!
1) Give the domain, range and zeros of the following graph:
2) Give the domain, range and zeros of the function: f(x) = 5x/(3x - 2)
3) Sketch the graph and use the graph to find the range and zeros of the function: f(x) = x3 + 4x2 - x - 4
4) Let f(x) = 4x, g(x) = x2, h(x) = x + 5 and j(x) = 2x. Find each.
a) (f + h)(x) b) (j - f)(x) c) f(g(h(x))) d) g(h(j(x))) e) (f/g)(x) f) g(f(h(2)))
5) Sketch the graphs of y = x2 -4, y = 4 - x2 and y = |4 - x| on the same axes.
6) Test each equation for all 4 symmetries.
a) x2 - xy = 3 b) x2 - y2 = 1 c) y = -|x|
7) Use symmetry to sketch the graph of xy2 = 2
8) Sketch each of the following by using
2
stretching/shrinking and/or translation.
a) y - 5 = 2|x - 1| b) y + 3 = -(x + 1)2
9) Prove that f(x) = (x+1)/2 and g(x) = 2x - 1 are inverse functions.
10) Sketch the graph of f and f -1. Then find a rule for the inverse. f(x) = x2 - 2, x < 0
11) a) Let P(x, y) = 2x + 2y. Let y be constant at y = 2, y = 3, y = 4. Graph these functions.
b) Now let P be constant at P = 10, P = 12 and P = 14. Graph these functions.
12) The area of a triangle is a function of the base b and height h.
a) Express A as a function of b and h. b) Find A(2,3) and A(5,6)
c) Draw the curve of constant area A(b, h) = 5
in a bh-plane.
13) let f(x) be the following graph:
a) What is the fundamental period for the above graph?
b) What is f(1)? f(2)? f(3)? f(4)? c) What is
f(99)? f(13)? f(-2)? f(-11)? d) Graph 2f(x) e) Graph f(2x) f) Graph -f(x) g) Graph f(-x)
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Answers To Sample Test
Domain -- x < 4 Range -- y > 0 Zero at ( 4, 0)
2) Domain -- all real numbers but 2/3 Range -- all real numbers but 5/3
Zero at ( 0, 0)
3) Range is all real numbers zeros at (-4, 0), (-1, 0), and (1, 0)
1)
4) a) 5x + 5 b) -2x
c) 4(x + 5)2 d) (2x + 5)2
e) 4/x, x can't equal 0 f) g[f(7)] = g(28) = 784
5) a) b) 6)
7) It is symmetric to the x-axis
c)
8) a) is a vertical stretch and translation of 1 right and 5 up
b) is a reflection about the x-axis and a translation of 1 left and 3 down
9) To prove this you must show both ways!! f[g(x)] = f(2x-1) = (2x - 1 + 1)/2 = 2x/2 = x
g[f(x)] = g[(x + 1)/2] = [2(x + 1)/2] - 1 = x + 1 - 1 = x
10) To find the rule interchange x and y and solve for y.
x = y2 - 2 x + 2 = y2
_____
y = - \\/ x + 2 , x > -2
11) a) b)
12) a) A(b, h) = bh/2
b) A(2, 3) = (2)(3)/2 = 3 and A(5, 6) = (5)(6)/2 = 15
c) 5 = bh/2 ==> 10 = bh ==> h = 10/b
13) a) 4 b) 1, 0, -1, 0 c) -1, 1, 0, 1
d) Changes amplitude but not period length.
e) Changes period length to 2 (horizontal shrink) but does not change amplitude
f) Reflection about the x-axis
g) Reflection about the y-axis
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Hope you did well on this opportunity. If not go back
and study the previous sections until you have mastered the material. Especially study the reflections, stretches and shrinks, and of course, the
translations. Good luck!!
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