Combinations
of Functions – Click here for logarithmic and exponential functions
In a Nut Shell: Functions of one (or more)
independent variables can be combined in various ways to develop
new functions. In each of the simple cases
below the single independent variable is x. (cf)(x) = c
f(x) constant, c, times a
function, f(x) (f + g)(x)
= f(x) + g(x) addition of functions f(x) and g(x) (f – g)(x)
= f(x) – g(x) subtraction of g(x)
from f(x) (f * g)(x)
= f(x) * g(x) where
* denotes multiplication
of f(x) times g(x) (f / g)(x)
= f(x) / g(x) division of f(x) / g(x) where
g(x) ≠ 0 |
Here are four simple examples using f(x)
= 3x + 5 and g(x)
= √6x (f + g)(x) = 3x
+ 5 + √6x obtain new function by addition (f – g)(x) = 3x
+ 5 - √6x obtain new function by
subtraction (f * g)(x) =
(3x + 5) (√6x)
obtain new function by multiplication (f/g)(x) =
(3x + 5) / (√6x) obtain new function by division |
Another function, called the composite function, can be formed by
replacing the independent variable, x, in one function with the other function. For
example: Let f(x) and
g(x) be defined as follows: f(x)
= 3x + 5, g(x) = √6x The composition of f and g
is expressed as f( g(x) ), reads as
f of g(x). That is replace
x with g(x). In this case f(g(x) ) = 3(√6x)
+ 5 Likewise the composition
of g
and f is
g( f(x) ) , reads as g of
f(x). That is replace
x with f(x). In this case g(f(x)
) =
√ [ 6(3x + 5) ] Note that this type of
calculation will be used later for the chain rule of differentiation of functions. The chain rule is very important. Click to continue with
discussion of logarithmic and exponential functions. |
Return to Notes for Calculus 1 |
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