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1.3 Functions

1.3 Functions. Domain and Range Function notation Piecewise functions Interval Notation Difference quotient. Domain and Range. Domain is the Input (Independent variable) used in a function. Range is the Output (Dependent variable) given by the function.

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1.3 Functions

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  1. 1.3 Functions Domain and Range Function notation Piecewise functions Interval Notation Difference quotient

  2. Domain and Range Domain is the Input (Independent variable) used in a function. Range is the Output (Dependent variable) given by the function. Relations are functions, functions match elements from the domain to the range

  3. A Function A car’s brake pedal is part of a function car. Step on the brake, the car should stop. Every time.

  4. Elements in the Domain only map to one element in the Range a x An element in the b y Range can have c z many Inputs, Where d w the Domain can only go to one element in the Range.

  5. Ways to Represent a function Verbally How the input effect the output Numerically A table of numbers Graphically With a graph Algebraically With an equation

  6. Testing a Function Verbally with a Proof Numerically with a table. If an element from the Domain has two different outcomes, it is not a function. For example (1, 5), (2, 3) and (2,1).

  7. Testing a function Graphically uses the vertical line test. If a vertical line touches the graph in more then one spot, then the graph is not the graph of a function. Algebraically is where you solve an element of the range. X2+ y= 8 Solve for y= - x2+8 Y has only one answer so it is ok x2+y2=25 Solve for Not a function

  8. Function notation Use f(x) to stand in for y in the equation y = 5x – 2 , so it becomes f(x) = 5x – 2 Why would we need this notation?

  9. Solve a function with different inputs Let k(x) = 2x2-1 So find k(x) for x ={0, 1, 2} k(0) = 2(0)2- 1 = -1 k(1) = 2(1)2- 1 = 1 k(2) = 2(2)2- 1 = 7

  10. Sometime the input can be an expression h(x) = x2 – 4x + 6 Let x = 2 h(2) = (2)2 – 4(2) + 6 = 2 Let x = (x + 1) h(x+1) = (x+1)2 – 4(x+1) + 6 h(x+1) = (x2 2x + 1) – 4x – 4 + 6 h(x+1) = x2 – 2x + 3

  11. Piecewise functionsdifferent functions over different parts of a domain Here is the Rule of the piecewise function (0,2) If x = - 2 , then 2(-2) + 2 = -2 If x = 0 , then -3(0) – 1 = -1 If x = 1, then -3(2) – 1 = -4 (0, - 1) (1, -4) (-2, -2)

  12. Implied Domain, the real numbers in which the function is defined Domain where x does not equal 5 or All real numbers except 5 All real numbers except Where x greater then or equal to 0

  13. Ways to find the Implied Domains Find what makes the denominator of a fraction zero. Find what makes an even root (square root, 4th roots and so on) inside negative.

  14. Interval Notation Find the Implied Domain After setting 9 – x2 = 0 and finding the answer -3 ≤ x ≤ 3 , easier said then done. Since x includes -3 and 3 we use [ ] to show the number between So -3 ≤ x ≤ 3 , becomes [-3, 3] If it was 4 < y ≤ 10, it would be (4, 10] ( ) shows that the end numbers are not included in the solution set.

  15. The Difference Quotient A f(x) = x2+ 3x + 5 In the Difference Quotient (x+h)2+3(x+h)+5-(x2+3x+5) h

  16. = x2+2hx+h2+3x+3h+5-x2-3x-5 h = 2hx + h2+3h = h(2x + h + 3) h h = 2x + h + 3 h not equaling 0

  17. Homework Page 35 – 38 #4, 7, 8, 10, 16, 20, 22, 26, 34, 36, 40, 46, 48, 50, 54, 60, 66, 71, 74, 80, 86, 90

  18. Homework Page 36-39 # 35, 49, 53, 61, 65, 79, 87, 91, 96, 98, 101,103, 104,105

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