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f. x. y =f( x ). Functions A function is an operation performed on an input (x) to produce an output (y = f(x) ). The Domain of f is the set of all allowable inputs (x values) The Range of f is the set of all outputs (y values). Domain. Range. To be well defined a function must

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## Functions A function is an operation performed on an input (x) to produce an output (y = f(x) ).

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**f**x y =f(x) Functions A function is an operation performed on an input (x) to produce an output (y = f(x) ). The Domain of f is the set of all allowable inputs (x values) The Range of f is the set of all outputs (y values) Domain Range**To be well defined a function must**· Have a value for each x in the domain · Have only one value for each x in the domain e.g y = f(x) = √(x-1), x is not well defined as if x < 1 we will be trying to square root a negative number. y = f(x) = 1/(x-2), x is not well defined as if x = 2 we will be trying to divide by zero. This is not a function as some x values correspond to two y values.**Finding the Range of a function**Draw a graph of the function for its given Domain The Range is the set of values on the y-axis for which a horizontal line drawn through that point would cut the graph. The Function is f(x) = (x-2)2 +3 , x y = (x-2)2 +3 The Range is f(x) ≥ 3 Domain y = (x-2)2 +3 2 Range 3 Link to Inverse Functions Domain**The Function is f(x) = 3 – 2x , x**The Range is f(x) < 3**Composite Functions**Finding gf(x) Note: gf(x) does not mean g(x) times f(x). g f f(x) x g(f(x)) = gf(x) Note : When finding f(g(x)) Replace all the x’s in the rule for the f funcion with the expression for g(x) in a bracket. e.g If f(x) = x2 –2x then f(x-2) = (x-2)2 – 2(x-2) gf(x) means “g of f of x” i.e g(f(x)) . First we apply the f function. Then the output of the f function becomes the input for the g function. Notice that gf means f first and then g. Example if f(x) = x + 3, x and g(x) = x2 , x then gf(x) = g(f(x)) = g(x + 3) = (x+3)2 , x fg(x) = f(g(x)) = f(x2) = x2 +3, x g2(x) means g(g(x)) = g(x2) = (x2)2 = x4 , x f2(x) means f(f(x)) = f(x+3) = (x+3) + 3 = x + 6 , x**Notice that fg and gf are not the same.**The Domain of gf is the same as the Domain of f since f is the first function to be applied. The Domain of fg is the same as the Domain of g. For gf to be properly defined the Range (output set) of f must fit inside the Domain (input set) of g. For example if g(x) = √x , x ≥ 0 and f(x) = x – 2, x Then gf would not be well defined as the output of f could be a negative number and this is not allowed as an input for g. However fg is well defined, fg(x) = √x – 2, x ≥ 0.**f**a b f-1 Inverse Functions. The inverse of a function f is denoted by f-1 . The inverse reverses the original function. So if f(a) = b then f-1(b) = a Note: f-1(x) does not mean 1/f(x). Domain of f Range of f = Domain of f-1 = Range of f-1**One to one Functions**If a function is to have an inverse which is also a function then it must be one to one. This means that a horizontal line will never cut the graph more than once. i.e we cannot have f(a) = f(b) if a ≠ b, Two different inputs (x values) are not allowed to give the same output (y value). For instance f(-2) = f(2) = 4 y = f(x) = x2 with domain x is not one to one. So the inverse of 4 would have two possibilities : -2 or 2. This means that the inverse is not a function. We say that the inverse function of f does not exist. If the Domain is restricted to x ≥ 0 Then the function would be one to one and its inverse would be f-1(x) = √x , x ≥ 0**Finding the Rule and Domain of an inverse function**• Rule • Swap over x and y • Make y the subject Domain The domain of the inverse = the Range of the original. So draw a graph of y = f(x) and use it to find the Range Drawing the graph of the Inverse The graph of y = f-1(x) is the reflection in y = x of the graph of y = f(x).**Example:**Find the inverse of the function y = f(x) = (x-2)2 + 3 , x ≥ 2 Sketch the graphs of y = f(x) and y = f-1(x) on the same axes showing the relationship between them. Domain This is the function we considered earlier except that its domain has been restricted to x ≥ 2 in order to make it one-to-one. We know that the Range of f is y ≥ 3 and so the domain of f-1 will be x ≥ 3. Rule Swap x and y to get x = (y-2)2 + 3 Now make y the subject x – 3 = (y-2)2 √(x –3) = y-2 y = 2 + √(x –3) So Final Answer is: f-1(x) = 2 + √(x –3) , x ≥ 3 Graphs Reflect in y = x to get the graph of the inverse function. Note: we could also have -√(x –3) = y-2 and y = 2 - √(x –3) But this would not fit our function as y must be greater than 2 (see graph) Note: Remember with inverse functions everything swaps over. Input and output (x and y) swap over Domain and Range swap over Reflecting in y = x swaps over the coordinates of a point so (a,b) on one graph becomes (b,a) on the other.

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