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HW 1.4: Pg. 48-50 #9, 30, 31, 35, 57-69 odd Quiz next class!. September 12, 2012 - Determine whether relationships between two variables are functions - Use function notation and evaluate functions - Find the domains of functions. Warm-up: Copy Quiz Topics Linear Equations + word problems - PowerPoint PPT Presentation
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September 12, 2012- Determine whether relationships between two variables are functions- Use function notation and evaluate functions- Find the domains of functions
Warm-up: Copy Quiz Topics• Linear Equations + word problems• Circle equation• x- and y-intercepts• Symmetry• Functions – evaluating
HW 1.4: Pg. 48-50 #9, 30, 31, 35, 57-69 odd
Quiz next class!
Check 1.2-1.3 Word Problems
Pg. 23 #72 Compare graphs
Pg. 36 #96 a) Largest increase 2002-03; least increase 2000-2001 b) slope from 94-2003 = 9.18 c)
#106 y(t) = -749.5t + 10,821.5; y(18) = -2669.5;
y(20) = -4168.5
#109 a) y(t) = 179.5t + 40,571; b) y(8) = 42,007;
y(10) = 42,366; c) m = 179.5
Lesson 1.4 Functions
A function f from a set A to a set B is a relation that assigns to each element x in the set A exactly one element y in the set B. The set A is the domain (or set of inputs) of the function f, and the set B contains the range (or set of outputs).
• Verbally • Numerically • Graphically• Algebraically
“The input value x is the number of representatives from a state, and the output value of y is the number of senators.”
Four Ways to Represent a Function
(2, 11), (2, 10), (3, 8), (4, 5)
y = x2 + 2x + 1
Function Notation f(x)
Ex 1) Evaluating a function g(x) = -x2 + 4x + 1
Find the function value for the following:
a) g(2) b) g(t) c) g(x + 2)
g(2) = -(2)2 + 4(2) + 1
Replace all the x’s with the value of 2
g(2) = -4+ 8 + 1
g(2) = 5
g(t) = -(t)2 + 4(t) + 1
g(t) = -t2 + 4t + 1
g(x +2) = -(x+2)2 + 4(x + 2) + 1
g(x + 2) = -(x+2)(x+2) + 4(x + 2) + 1
g(x + 2) = -(x2 + 4x + 4)+ 4(x + 2) + 1
g(x + 2) = -x2 - 4x - 4+ 4x + 8 + 1
g(x +2) = -x2 + 5
Evaluating a piecewise-defined function (two or more equations over a specified domain)
Example 2) Evaluate the function when x = -1, 0, and 1
0,1
0,1)(
2
xx
xxxf
It gives you two conditions:
- Use x2 + 1, if x < 0
- Use x – 1, x ≥ 0
f(-1) = (-1)2 + 1
f(-1) = 1 + 1
f(-1) = 2
f(0) = 0 – 1
f(0) = – 1
f(1) = 1 – 1
f(1) = 0
Practice
1. Evaluate f(x) = x2 + 1 for:
a) f(-4) b) f(t2) c) f(t + 1)
2. Evaluate for:
a) f(-2) b) f(-1) c) f(0)
1,1
1,12)(
2 xx
xxxf
17 t4 + 1 t2 + 2t + 2
-3 -1 1
Finding the Domain of a Function
There are exceptions to what values can be inputted into a function.
For example:
A fraction A square root Quadratic
f (x) 1
x
f (x) x
{x| x is all real numbers, x ≠ 0} {x| x ≥ 0} {x| x is all real numbers}
Graphing the function also helps you see the domain.
f (x) x 2
PracticeFind the domain for each function
3. 4. 5.
5
1)(
x
xg 24)( xxf 13)( xxf
Evaluating a Difference QuotientThe difference quotient is the slope of the secant line that goes through two points of a function. This slope is very important in calculus where it is used
to define the derivative of function f.
h
xfhxf )()(
Answer 2x + h – 4
Evaluate f(x) = x2 – 4x + 7