College Algebra

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Previous Lessons
Open Chapter Ch. 1: Graphs, Functions, and Models
Lesson #1 Graphing
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Lesson #2 Graphing Functions
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Lesson #3 Slope of Linear Functions
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Lesson #4 Linear Equations and Modeling
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Lesson #5 Linear Equations, Functions, Zeros and Applications
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Lesson #6 Linear Inequalities
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Open Chapter Ch. 2: More on Functions
Lesson #7 Analyzing Functions
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Exam Exam 1
Lesson #8 Algebra for Functions
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Lesson #9 The Composition of Functions
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Lesson #10 Symmetry
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Lesson #11 Transformations
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Lesson #12 Equations of Variation
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Open Chapter Ch. 3: Quadratic Functions and Equations; Inequalities
Lesson #13 Complex Numbers
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Lesson #14 Quadratic Equations, Functions, Zeros and Models
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Lesson #15 Analyzing Graphs of Quadratic Functions
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Lesson #16 Rational and Radical Equations
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Exam Midterm Exam
Lesson #17 Absolute Value Equations and Inequalities
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Open Chapter Ch. 4: Polynomial Functions and Rational Functions
Lesson #18 Polynomial Functions and Models
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Lesson #19 Graphing Polynomial Functions
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Lesson #20 Polynomial Division
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Lesson #21 The Zeros of Polynomial Functions
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Lesson #22 Rational Functions
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Lesson #23 Polynomial and Rational Inequalities
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Open Chapter Ch. 5: Exponential Functions and Logarithmic Functions
Lesson #24 Inverse Functions
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Lesson #25 Exponential Functions and Graphs
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Lesson #26 Logarithmic Functions and Graphs
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Exam Exam 3
Lesson #27 Properties of Logarithmic Functions
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Lesson #28 Solving Exponential Equations and Logarithmic Equations
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Lesson #29 Applications and Models: Growth and Decay; Compound Interest
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Open Chapter Ch. 6: Systems of Equations
Lesson #30 Systems of Equations in Two Variables
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Lesson #31 Systems of Inequalities and Linear Programming
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Lesson #32 Nonlinear Systems of Equations and Inequalities
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Lesson #33 Sequences and Series
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Exam Final Exam

Assignments:

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Lesson Objectives:

- Find the inverse of a one-to-one function.
- Simplify the composition of a function and its inverse.


Given a relation defined by a set of ordered pairs, the inverse relation is obtained by interchanging the first and second coordinates of each ordered pair.

If the relation is defined by an equation, then the inverse relatino is obtained by interchanging the values.



Find the inverse of the relation.

`{(6, 7), (-3,7), (2, -5), (7, -9)}`.

We have to interchange the first and second coordinates of ach ordered pair in order to produce the inverse relation.

So the inverse relation is the set of ordered pairs, (7,6), (7, -3), (-5, 2), and (-9, 7).



Find an equation of the inverse relation.

`x = y^2-3y`. Now we need to interchange the variables x and y in order to produce the inverse relation. So we have `y = x^2-3x`. Now let's graph both of these equations.

We'll label the first equation 1, and the second equation 2. So this is the graph of our first equation, and this is the graph of our second equation. This is the line, `y = x`. The graphs of a relation and its inverse are always reflections of each other over the line `y = x`.



One-to-One Functions

A function is one-to-one if the outputs are the same when the inputs are the same, and the outputs are different, when the inputs are different.

One-to-One Functions and Inverses

The inverse function is called "f inverse", with an `f` and a -1 in the corner, where -1 is not a power. So "f inverse" is not equivalent to `1/f`.

If `f` is a one-to-one function, then its inverse is also a function. And the domain of `f` is the range of its inverse. The range of `f` is the domain of its inverse.



Now prove that `f` is one-to-one, using the definition of a one-to-one function.

`f(x) = (1/4)x-6`. We'll need to show that `f(a) = f(b)`, when `a = b`. So, `f(a) = 1/4(a-6)`, and `f(b) = 1/4(b-6)`. So if we add 6 to both sides, that cancels, and we're left with `1/4a = 1/4b`. And then multiply both sides by 4 and we're left with `a = b`.

So if `f(a) = f(b)`, then `a = b`. This proves that `f` is one-to-one.



Prove that `g` is not one-to-one, using the definition of a one-to-one function.

We're given `g(x) = 5-x^2`. Now we need to show that when `g(a) = g(b)`, `a != b`. Because of this `x^2` term here, we know that if we let `a = -2` and `b = +2`; those are two terms which don't equal each other, or which g(a) and g(b) will be equal to each other. `5-(-2)^2` is the same as `5-(2)^2`. We get `1 = 1`.

So since `g(a) = g(b)`, when `a != b`, then g(x) is not one-to-one.



The Horizontal Line Test.

If a horizontal line can cross the graph of a function more than once, then the function is not one-to-one.

And remember that if a function is not one-to-one, then its inverse is not a function.



Using the horizontal line test, determine whether each function is one-to-one.

So for our first graph, we're given `f(x) = (7.2)^x`. Now if we draw a horizontal line, it strikes only once. So, yes, this is a one-to-one function.

Now for our second graph, we're given `f(x) = 5-x^2`. And if we draw a horizontal line here, it strikes the graph more than once. So, no, this is not a one-to-one function.

Since `f(x) = (7.2)^x` is one-to-one, we know that its inverse is a function.



For the function, determine if it is one-to-one, and if it is, find a formula for the inverse.

`f(x) = 3x-1`. This is of the form, `y = mx+b`. It's a straight line, so we know that the y-intercept occurs at (0, b) or (0, -1). And the slope, m, is equal to `3/1`.

So now if we draw the graph, we have our y-intercept at (0, -1), and our slope is 3 units up, 1 unit to the right. So now if we use our horizontal line test, we can see that a horizontal line, drawn anywhere on this graph, can only strike the graph once. So it is a one-to-one function.

Now let's find the formula for the inverse function. We'll start by replace f(x) with y. So we have `y = 3x-1`. And then interchange the variables, so we have `x = 3y-1`, and solve for y. So we have `x+1 = 3y`, when we add 1 to both sides. And then when we divide both sides by 3, we get `(x+1)/3 = y`. So we have `f^(-1)(x) = (x+1)/3`.



Given the graph of the one-to-one function, `f`, sketch the graph of f inverse.

So f contains the points: (-2, -7), (-1, 0), (0, 1), (1, 2), and (2, 9). So f inverse will contain the points: (-7, -2), (0, -1), (1, 0), (2, 1), and (9, 2).

This is the graph of f inverse. It's the reflection of the graph of f, across the line `y = x`.



Inverse Function and Composition

If `f` is a one-to-one function, then the composition of f inverse and f is equal to x, for every x in the domain of f. And the composition of f and f inverse is equal to x, for every x in the domain of f inverse.



For the function, `f`, use composition of functions to show that f inverse is as follows:

If `f(x) = (2/3)x+1` then `f^(-1)(x) = (3x-3)/2`. So first we need to find the composition of `f^(-1)` and f, or input f into `f^(-1)`. So we have `(3(2/3x+1)-3)/2`. We can simplify this as `(2x+3-3)/2`, and 3-3 cancels, and we're left with `2x/2`. And then the 2's cancel, so all of this is equal to x.

And now we need to find the composition of f and `f^(-1)`. This is the same as inputting `f^(-1)` into f. So we have `f((3x-3)/2)`, which is equal to `2/3((3x-3)/2)+1`. 2's cancel, the 3's cancel, and we're left with `x-1+1`, which equals x.

So since the composition of `f^(-1)` and f, and the composition of f and `f^(-1)` both equal x, that means that `f^(-1)` is as given.