- Ilock 3 0 2 X 2
- Ilock 3 0 2 X 2 1 4 Inner Tube
- Ilock 3 0 2 X 2 1 2 Square Tubing
- Ilock 3 0 2 X 2 0 For X
- Ilock 3 0 2 X 2 Aluminum Square Tube
Finding Domain and Range
Function | Notes |
If x = 0, you would be dividing by 0, so x ≠ 0. | |
If x = 3, you would be dividing by 0, so x ≠ 3. | |
Although you can simplify this function to f(x) = 2, when x = 1 the original function would include division by 0. So x ≠ 1. | |
Both x = 1 and x = −1 would make the denominator 0. Again, this function can be simplified to , but when x = 1 or x = −1 the original function would include division by 0, so x ≠ 1 and x ≠ −1. | |
This is an example with no domainrestrictions, even though there is a variable in the denominator. Since x2 ≥ 0, x2 + 1 can never be 0. The least it can be is 1, so there is no danger of division by 0. |
Function | Restrictions to the Domain |
If x < 0, you would be taking the square root of a negative number, so x ≥ 0. | |
If x < −10, you would be taking the square root of a negative number, so x≥ −10. | |
When is -x negative? Only when x is positive. (For example, if x = −3, then −x = 3. If x = 1, then −x = −1.) This means x ≤ 0. | |
x2 – 1 must be positive, x2 – 1 > 0. So x2 > 1. This happens only when x is greater than 1 or less than −1: x ≤ −1 or x ≥ 1. | |
There are no domainrestrictions, even though there is a variable under the radical. Since x2 ≥ 0, x2 + 10 can never be negative. The least it can be is 10, so there is no danger of taking the square root of a negative number. |
Domains can be restricted if: ·the function is a rational function and the denominator is 0 for some value or values of x. ·the function is a radical function with an even index (such as a square root), and the radicand can be negative for some value or values of x. |
Quadratic function,f(x) = x2 – 2x – 3 |
Rational function,f(x) = |
Example | |
Problem | What are the domain and range of the real-valued function f(x) = x + 3? |
This is a linear function. Remember that linear functions are lines that continue forever in each direction. Any real number can be substituted for x and get a meaningful output. For any real number, you can always find an x value that gives you that number for the output. Unless a linear function is a constant, such as f(x) = 2, there is no restriction on the range. | |
Answer | The domain and range are all real numbers. |
Example | |
Problem | What are the domain and range of the real-valued function f(x) = −3x2 + 6x + 1? |
This is a quadratic function. There are no rational or radical expressions, so there is nothing that will restrict the domain. Any real number can be used for x to get a meaningful output. Because the coefficient of x2 is negative, it will open downward. With quadratic functions, remember that there is either a maximum (greatest) value, or a minimum (least) value. In this case, there is a maximum value. The vertex, or turning point, is at (1, 4). From the graph, you can see that f(x) ≤ 4. | |
Answer | The domain is all real numbers, and the range is all real numbers f(x) such that f(x) ≤ 4. |
Example | |
Problem | What are the domain and range of the real-valued function ? |
This is a radical function. The domain of a radical function is any x value for which the radicand (the value under the radical sign) is not negative. That means x + 5 ≥ 0, so x ≥ −5. Since the square root must always be positive or 0, . That means . | |
Answer | The domain is all real numbers x where x ≥ −5, and the range is all real numbers f(x) such that f(x) ≥ −2. |
Example | |
Problem | What are the domain and range of the real-valued function ? |
This is a rational function. The domain of a rational function is restricted where the denominator is 0. In this case, x + 2 is the denominator, and this is 0 only when x = −2. For the range, create a graph using a graphing utility and look for asymptotes: One asymptote, a vertical asymptote, is at x =−2, as you should expect from the domain restriction. The other, a horizontal asymptote, appears to be around y = 3. (In fact, it is indeed y = 3.) | |
Answer | The domain is all real numbers except −2, and the range is all real numbers except 3. |
Find the domain and range of the real-valued function f(x) = x2+ 7. A) The domain is all real numbers and the range is all real numbers f(x) such that f(x) ≥ 7. B) The domain is all real numbers x such that x ≥ 0 and the range is all real numbers f(x) such that f(x) ≥ 7. C) The domain is all real numbers x such that x ≥ 0 and the range is all real numbers. D) The domain and range are all real numbers. |
Graphing Linear Inequalities
Ilock 3 0 2 X 2
Ilock 3 0 2 X 2 1 4 Inner Tube
Ordered Pair | Makes the inequality 3 x + 2y≤ 6 a true statement | Makes the inequality 3 x + 2y≤ 6 a false statement |
(−5, 5) | 3(−5) + 2(5) ≤ 6 −15 +10 ≤ 6 −5 ≤ 6 | |
(−2, −2) | 3(−2) + 2(–2) ≤ 6 −6 + (−4) ≤ 6 –10 ≤ 6 | |
(2, 3) | 3(2) + 2(3) ≤ 6 6 + 6 ≤ 6 12 ≤ 6 | |
(2, 0) | 3(2) + 2(0) ≤ 6 6 + 0 ≤ 6 6 ≤ 6 | |
(4, −1) | 3(4) + 2(−1) ≤ 6 12 + (−2) ≤ 6 10 ≤ 6 |
Example | ||
Problem | Use the graph to determine which ordered pairs plotted below are solutions of the inequality x – y < 3. | |
Solutions will be located in the shaded region. Since this is a “less than” problem, ordered pairs on the boundary line are not included in the solution set. | ||
(−1, 1) (−2, −2) | These values are located in the shaded region, so are solutions. (When substituted into the inequality x – y < 3, they produce true statements.) | |
(1, −2) (3, −2) (4, 0) | These values are not located in the shaded region, so are not solutions. (When substituted into the inequality x – y < 3, they produce false statements.) | |
Answer | (−1, 1), (−2, −2) |
Example | ||
Problem | Is (2, −3) a solution of the inequality | |
y < −3x + 1 | If (2, −3) is a solution, then it will yield a true statement when substituted into the inequality y < −3x + 1. | |
−3 < −3(2) + 1 | Substitute x = 2 and y = −3 into inequality. | |
−3 < −6 + 1 | Evaluate. | |
−3 < −5 | This statement is not true, so the ordered pair (2, −3) is not a solution. | |
Answer | (2, −3) is not a solution. |
Which ordered pair is a solution of the inequality 2y - 5x < 2? A) (−5, 1) B) (−3, 3) C) (1, 5) D) (3, 3) |
![Ilock Ilock](https://ilock.com.ua/media/catalog/product/cache/1/image/9df78eab33525d08d6e5fb8d27136e95/m/u/mul-t-lock-classic-kt-lat-f-2_51.jpg)
Ilock 3 0 2 X 2 1 2 Square Tubing
Graphing Inequalities To graph an inequality: oGraph the related boundary line. Replace the <, >, ≤ or ≥ sign in the inequality with = to find the equation of the boundary line. oIdentify at least one ordered pair on either side of the boundary line and substitute those (x, y) values into the inequality. Shade the region that contains the ordered pairs that make the inequality a true statement. oIf points on the boundary line are solutions, then use a solid line for drawing the boundary line. This will happen for ≤ or ≥ inequalities. oIf points on the boundary line aren’t solutions, then use a dotted line for the boundary line. This will happen for < or > inequalities. |
−1 + 4(3) ≤ 4 |
−1 + 12 ≤ 4 |
11 ≤ 4 |
2 + 4(0) ≤ 4 |
2 + 0 ≤ 4 |
2 ≤ 4 |
Example | |||||||
Problem | Graph the inequality 2y > 4x – 6. | ||||||
Solve for y. | |||||||
| Create a table of values to find two points on the line, or graph it based on the slope-intercept method, the b value of the y-intercept is -3 and the slope is 2. Plot the points, and graph the line. The line is dotted because the sign in the inequality is >, not ≥ and therefore points on the line are not solutions to the inequality. | ||||||
2y > 4x – 6 Test 1: (−3, 1) 2(1) > 4(−3) – 6 2 > –12 – 6 2 > −18 True! Test 2: (4, 1) 2(1) > 4(4) – 6 2 > 16 – 6 2 > 10 False! | Find an ordered pair on either side of the boundary line. Insert the x- and y-values into the inequality 2y > 4x – 6 and see which ordered pair results in a true statement. Since (−3, 1) results in a true statement, the region that includes (−3, 1) should be shaded. | ||||||
Answer | The graph of the inequality 2y > 4x – 6 is: |
Ilock 3 0 2 X 2 0 For X
When plotted on a coordinate plane, what does the graph of y ≥ x look like? A) B) C) D) |