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 sum, difference, product and quotient of two functions. Then determine the resulting domain.
- Find the difference quotient for a function.


Adding, subtracting, multiplying and dividing functions.

In order to add `(f+g)(x)`, you have to add `f(x) + g(x)`. Then, in order to subtract `(f-g)(x)`, you have to subtract `f(x) - g(x)`. And then to multiply `(fg)(x)`, you have to multiply `f(x)*g(x)`. And then `(f/g)(x)` is the same as `f(x)/g(x)`, where `g(x)` cannot equal zero.



Given that `h(x) = x+3` and `g(x) = sqrt(x-4)`, what is `(g+h)(4)`?

`(g+h)(4)` is the same as `g(4)+h(4)`. So we first need to find what `g(4)` is and then find what `h(4)` is. So, `g(4)` is the `sqrt(4-4)` or the `sqrt(0)`, which is 0. And `h(4)` is `4+3`, which is 7. `(g+h)(4)` is `0+7`, which is 7.



The domain of f+g, f-g, `f*g`, and f/g is the intersection of the domain of f and the domain of g.

With the exception for f/g that it must not include x values which cause the denominator, `g(x) = 0`.



Given the pair of functions, `f(x) = 4x-1` and `g(x) = -4x^2`, what's the domain of f, g, f+g, f-g, `f*g`, `f*f`, f/g and g/f?

The domain of f is all real numbers, and the domain of g is all real numbers. This is written as `(-oo, oo)` in interval notation.

Therefore the domain of f+g, f-g, `f*g`, and `f*f` are the intersection of `(-oo, oo)` and `(-oo, oo)`, which is `(-oo, oo)`.

When finding the domain of f/g and g/f, you have to make sure that we don't have a zero in the denominator. So for f/g, the domain does not include `x = 0`, since `g(0) = 0`. The domain of f/g is `(-oo, 0) cup (0, oo)`.

Now for g/f, the domain does not include `x = 1/4`, since `4(1/4)-1` is zero, and that would give us zero in the denominator. The domain of g/f is `(-oo, 1/4) cup (1/4, oo)`.



Given the pair of functions, `f(x) = 4x-1` and `g(x) = -4x^2`, the same functions as before, find `(f+g)(x), (f-g)(x), (fg)(x), (ff)(x), (f/g)(x), "and" (g/f)(x)`.

So `(f+g)(x)` is `f(x)+g(x)`, or `-4x^2+4x-1`.
Then `(f-g)(x)`, `f(x)-g(x)`, which is `4x^2+4x-1`.
Then `(fg)(x)` is `(4x-1)(-4x^2)`, which is `-16x^3+4x^2`.
Then `(ff)(x)` is `f(x)*f(x)`, which is `16x^2-4x-4x+1`, and this can be simplified as `16x^2-8x+1`.
Now `(f/g)(x)` is the same as `f(x)/g(x)`.
Then `(g/f)(x)` is the same as `g(x)/f(x)`.



The slope of a line is an average rate of change. Now, consider the nonlinear function, f, with the secant line drawn between two points on the curve. Our two points are (x, f(x)) and (x+h, f(x+h)).

Let's find the slope of the secant line. m is going to equal the difference of y's, `f(x+h)-f(x)` over the difference in x's, `x+h-x`. And since the x-values in the denominator cancel, we're left with the slope equals `(f(x+h)-f(x))/h`. The slope of the secant line is called the difference quotient or the average rate of change.



The following functions are for total revenue, R(x), and total cost, C(x), of adding a DVD to Joe's Movie Rental Box.

Given the function, `R(x) = 90x-0.6x^2` and `C(x) = 5x+12`, the total profit can be found by subtracting cost from revenue. Find P(x). P(x) is revenue minus the cost, `90x-0.6x^2-(5x+12)`. So that can simplify to `90x-0.6x^2-5x-12`, which can be simplified further to `-0.6x^2+85x-12`.

So the total profit in adding 100 DVDs to Joe's Movie Rental box is P(100). P(100) is `-0.6(100)^2+85(100)-12`, which is 2488.



What is the difference quotient for the following function, `f(x) = 1/(2x)`.

The difference quotient is `(f(x+h)-f(x))/h`. So given f(x), we have to find f(x+h), so it's `1/(2(x+h))`. So the difference quotient is `(1/(2(x+h))-1/(2x))/h`. So let's create common denominators from multiplying our first fraction by 2x on the top and bottom, and `2(x+h)` on our second fraction. So the 2's cancel, and we can subtract the fractions to get `((x-x-h)/(2x(x+h)))/h`. Then the x's cancel, and `-h/h` is -1, and we're left with `-1/(2x(x+h))`.