1.4 Building Functions from Functions: * Functions can be combined algebraically using addition, subtraction, multiplication, and division to create new functions. EX: Find each given f(x) = 2x + 3 and g(x) = -­‐3x a) ( f + g)(x) b) ( f − g)(x) c) ( fg)(x) ⎛ f⎞
d) ⎜ ⎟ (x) ⎝ g⎠
EX: Let f (x) = x 2 and g(x) = x + 1 . Find the new function and give the domain for each. a) ( f + g)(x) b) ( f − g)(x) c) ( fg)(x) ⎛ f⎞
d) ⎜ ⎟ (x) ⎝ g⎠
Composition of Functions: • Function composition is the act of taking one function and substituting in for the variable in the other. EX: Let f (x) = x 2 − 1 and let g(x) = x . Find the composite function and determine its domain. a) g ! f b) f ! g Decomposing Functions: * When we decompose a function we reverse the process we just performed. We take a single function and break it down into two individual parts. To decompose a function, look for an “inner function” that you can remove. Typically you would find these functions inside parentheses, inside a radical, or in the denominator of a fraction. EX: ON YOUR OWN: h(x) = x 3 + 1 Relations and Implicitly Defined Functions: The general term for a set of ordered pairs (x, y) is a relation. If each x value only maps to one y value the relation is called a function and its graph will pass the vertical line test. When this is not the case and y is not a function of x, it is possible to split the equation into two equations that do represent functions. EX: Take the equation of the circle x 2 + y 2 = 4 . This graph would not pass the vertical line test and so is not a function. However if we were to solve for “y” we would be able to create two equations that do represent functions. ON YOUR OWN: Solve for “y” in the equation above to create two functions and graph each. Since all points in the two functions would satisfy the original equation, we say that the relation given by the equation defines the two functions implicitly.