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:

- Graph functions and determine which intervals are increasing, decreasing, or remaining constant. Estimate the relative maxima and minima.
- Find a function that models an application. Find the domain of a function and function values.
- Graph functions defined piecewise.


An increasing interval is one in which the graph of the function rises from left to right.

A decreasing interval is one in which the graph of a function drops from left to right.

And a constant interval is one in which the function values stay the same.



Determine the intervals on which the interval is increasing, decreasing, and constant.

As the x values increase from `-oo` to -1, the y values increase from `-oo` to 5. And as the x values increase from 5 to 8, the y-values increase from 0 to 7. Therefore, the interval of increasing is `(-oo, -1) and (5, 8)`. Notice that we use and and not unity, and notice that we're using parentheses.

As the x values are increasing from -1 to 5, the y values are decreasing from 5 to 0. Therefore, (-1, 5) is the decreasing interval.

And finally, as the x values are increasing from 8 to `oo`, the y values are remaining constant 7. So the constant interval is `(8, oo)`.



The relative maxima and minima.

f(c) is a relative maximum if (c, f(c)) is the highest point on some open interval. And f(c) is a relative minimum if the point (c, f(c)) is the lowest point on some open interval.



Using the graph, determine any relative maxima or minima of the function and the intervals on which the function is increasing or decreasing.

So here we're given the graph of `f(x) = 1/2x^3-1/4x^2-x+1`. The relative maxima can be found at the point (-0.67, 1.41) and it's the y-coordinate 1.41. And the relative minima can be found at the point (1, 0.25). It's the y-coordinate, 0.25, that is the relative minima.

So the relative maximum is 1.41 and the relative minimum is 0.25.



A Budweiser blimp is flying 3000 feet above the NRG stadium during the Super-Bowl. The slanted distance directly to the 50-yard line is d feet. Express the horizontal distance h as a function of d.

So here we have a right triangle, and we can use the Pythagorean theorem to setup our equation.

`h^2+3000^2 = d^2`

So since we want to express the horizontal distance as a function of d, let's solve for h. `h = sqrt(d^2-3000^2)`.

So we have the function, `h(d) = sqrt(d^2-3000^2)`.



A farmer has 350 feet of fencing with which to fence off three adjacent rectangular areas of grassland for his different livestock. A river forms one side of the fenced areas. Suppose the width of each area is x feet.

Now, express the total area of the three fenced off rectangles as a function of x. So, since x is our width, we have 4x for our combined widths. This means that 350 total feet of fencing - `4*x` will give us the length. So the total area as a function of x is going to equal `350-4x` times the width. So `A(x) = -4x^2+350x`.

Now, find the domain of the function. Since our smallest length is zero, we can solve for the smallest width by plugging in zero for length and solving for x. So x = 87.5. So the domain is (0, 87.5) in interval notation and `{x| 0 < x < 87.5}` in set notation. This is because the rectangle's length and width must be positive and there's only 350 feet of fencing.

Now use the graph of the function shown below to determine the dimensions that yield the maximum area. So since our function from part (a) was `A(x) = -4x^2+350x`, we can go ahead and label our axes A(x) and x. So looking at the graph, you can see that the maximum area occurs at the point (43.75, 7656.25), where A(x) is the y value; 7656.25 is the maximum area.

This occurs when the x value is 43.75 or the width is 43.75. So using this value, we can solve for the corresponding length. So the length for our maximum area is `350-4(43.75)`, which gives us 175. The dimensions that yield the maximum area is 43.75 feet by 175 feet.



For the piece-wise function, find the specified function values. We're given `g(x) = x+2 "for" x<=1` and `5-2 "for" x > 1` and we need to determine g(-5), g(0), g(1) and g(5).

First we determine which part of the domain contains the given input, and then we use the corresponding formula to find the output. So for g(-5), x = -5, which is less than or equal to 1. So we use `g(x) = x+2`. So this gives us `g(-5) = -5+2`. The output is -3.

For g(0), since x = 0, which is less than or equal to 1, we use the formula, `g(x) = x+2`, and we get an output of 2.

For g(1), x = 1, so x is less than or equal to 1, and we use the formula, `g(x) = x+2` again, and we get an output of 3.

And for g(5), we have x = 5, which is greater than 1, so we use the formula, `g(x) = 5-x`, and we get an output of 0.



Graph the following, `f(x) = 1/4x` when x is less than zero, and `x+5` when x is greater than or equal to zero.

So we graph `f(x) = 1/4x` on the interval `(-oo, 0)` and we graph `f(x) = x+5` on the interval `[0, oo)`. So, here are some ordered pairs where x is less than zero and x is greater than or equal to zero.

For `f(-8)`, we need to plug -8 in for x, over here, since -8 is less than zero. `1/4(-8)` gives us -2. So we have the point (-8, -2).

So for `f(-4)`, we need to plug -4 in for x, and `1/4(-4)` is -1. So our next ordered pair is (-4, -1).

Now for `f(0)`, we need to plug 0 in for x, and we get `0+5` is 5. This gives us the ordered pair (0, 5).

And now for `f(2)`, we plug 2 in for x, and we get `2+5` is 7. And we get the ordered pair (2, 7).

Now don't forget to plug zero into our first formula as well and have an open circle there. So `0*1/4` is zero and we'll have an open circle at the point (0, 0).

So for our first formula, we have this portion of the graph. Notice the open circle at (0, 0). And for our second formula, `x+5`, we have this portion of the graph.



Now there's such a thing called the greatest integer function.

`f(x) = [x]`. This means the greatest integer less than or equal to x. So for example, for the domain, if you have -4, -3.5, or -3.24, the range is always -4. And if in the domain you have the inputs: 2, 2.2, and 2.85, the range will always be 2.



Graph the following function: `f(x) = 2 + [x]`.

The greatest integer function can also be defined as a piece-wise function with an infinite number of statements. When x is greater than or equal to -3, but less than -2, the greatest integer value is -3. So, `-3+2 = -1`.

Then when x is greater than or equal to -2, but less than -1, the greatest integer value is -2. And `-2+2 = 0`.

Then when x is less than zero, but greater than or equal to -1. The greatest integer value is -1, and `-1+2 = 1`, etc.

So for such a function, we plot a closed point at (-3, -1), but an open point at (-2, -1). And then a closed point at (-2, 0), and then an open point at (-1, 0). And then a closed point at (-1, 1), and an open point at (0, 1). And then a closed point at (0, 2), and an open point at (1, 2), etc.

So here's a sample of our graph for `f(x) = 2 + [x]`.