# M5L1 Operations on Functions

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### M5L1 Operations on Functions adapted from hisema01:

M5L1 Operations on Functions adapted from hisema01 Objectives : Add, subtract, multiply, and divide functions. Compose functions Find the inverse of a function

### +, -, ×, ÷ Functions:

+, -, ×, ÷ Functions Sum : (f + g)(x) = f(x) + g(x) Difference : (f – g)(x) = f(x) – g(x) Product : (f ∙ g)(x) = f(x) ∙ g(x) Quotient :

### Example 1:

Example 1 Let f(x) = x + 1 and g(x) = x 2 – 1, find : Sum: ( f + g)(x ) = (x + 1)+( x 2 – 1) = x 2 + x Divide: =

### Example 2:

Example 2 Let f(x) = x 2 + x and g(x) = x + 1, find : Difference: (f – g)(x ) = (x 2 +x)-(x+1) = x 2 - 1 Product: (f ∙ g)(x ) = (x 2 + x)( x + 1 ) = x 3 + x 2 + x 2 + x = x 3 + 2x 2 + x

### Function Composition:

Function Composition One function becomes the input of another function (i.e. it’s plugged in for x) (f ◦ g)(x) = f(g(x)) g(x) becomes the input for f (g ◦ f)(x) = g(f(x)) f(x) becomes the input for g

### Example 3 ~ composition:

Example 3 ~ composition Let and Find ( f ◦ g)(x ) ~ plug g(x) into f(x) in place of x *The domain is still limited to because in the first step  numbers smaller than 2 would cause a negative inside the square root.

### Example 4 ~ composition:

Example 4 ~ composition Let and Find ( f ◦ g)(x) and (g ◦ f)(x ).

### Inverse of a function:

Inverse of a function To find the inverse of a function, you need to switch x & y, and solve for y. Example 5: f(x) = 6 x – 2 Remember f(x) = y So, y = 6x – 2 Switch x & y x = 6y – 2 Now solve for y x + 2 = 6y Addition Property Division Property The inverse is which is also a function.

### Inverse Function Notation:

Inverse Function Notation If the inverse of a function is also a function, we use special notation So for the last example since y = is a function, we can rewrite it as .