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1.4 BUILDING FUNCTIONS FROM FUNCTIONS By: Alaina Riedel & Lauren Evenson.

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1.4 BUILDING FUNCTIONS FROM FUNCTIONS By: Alaina Riedel & Lauren Evenson
Transcript

1.4 BUILDING FUNCTIONS FROM FUNCTIONS

By: Alaina Riedel & Lauren Evenson

“Building Blocks, Building Functions”

f(x) = 2x -

1

g(x) =

g(x) =

f(x) =

g(x) =

f(x) =

|x +

3|

g(x) = 3 -

xf(x) =

f(x) =

g(x) = 9 -

g(x) =

f(x) =

g(x) =

f(x) =

f(x) = 2

g(x) = (x

-

1

f(x) =

3x +

2

g(x) =

g(x) =

g(x) = f(x) = + 4

http://www.youtube.com/watch?v=9rsJF6lqxaoCopy and Paste into Internet Explorer

Combining Functions

A way of creating new functions to combine two or more functions to create a new function. The most obvious way we can do this is to

perform basic algebraic operations on the two functions to create the new one

Add, subtract, multiply or divide functions. The algebra of real numbers: 4 x 5 = 20,

4 - 5 = -1 The algebra of functions: fg, f-g, etc

Sum, Difference, Product, and Quotient

Composition of Functions

 

Relations and Implicitly Defined Functions

Relation: A set of ordered pairs of real numbers

- If the relation happens to relate a single value of y to each value of x, then the relation is also a function and its graph will pass the vertical line test

Implicitly defined function: A function that is a subset of a relation defined by an equation in x and y

Example - Relation

Determine which of the ordered pairs (2,-5), (1,3), and (2,1) are in the relation defined by + . Is this relation a function?

(2,-5): (2(5) + (-5 = 5(1,3): (1(3) + (3 = 12 = 5(2,1): (2(1) + (1 = 5So, (2,-5) and (2,1) are in the relation, but (1,3) is not. Since the equation relates two different y-values (-5 and 1) to the same x-value (2), the relation cannot be a function.

Example - Implicitly

The graph consists of two parallel lines, each the graph of one of the implicitly defined functions

jeopardylabs.com/play/building-functions-from-functions2

Alaina & Lauren Mix Up


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