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# Combinations of Functions Help

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## Introduction to Combinations of Functions

Most of the functions studied in calculus are some combination of only a few families of functions, most of the combinations are arithmetic. We can add two functions, f + g(x), subtract them, fg (x), multiply them, fg(x), and divide them . The domain of f + g ( x ), fg(x), and fg(x), is the intersection of the domain of f(x) and g(x). In other words, their domain is where the domain of f(x) overlaps the domain of g(x). The domain of is the same, except we need to remove any x that makes g(x) = 0.

#### Examples

Find f + g(x), fg(x), fg(x), and and their domain.

• f ( x ) = x 2 − 2 x + 5 and g (x) = 6 x −10

f + g (x) = f(x) + g (x) = (x2 − 2 x + 5) + (6 x − 10) = x 2 + 4 x − 5

f - g (x) = f(x) − g (x) = (x 2 − 2 x + 5) − (6 x − 10) = x 2 − 8x + 15

fg (x) = f(x) g (x) = (x2 − 2 x + 5)(6 x − 10) = 6 x 3 − 10 x2 − 12 x 2 + 20 x + 30 x − 50

= 6 x 3 − 22 x2 + 50 x − 50

The domain of f + g(x), fg (x), and fg(x) is (−∞, ∞). The domain of is (from 6 x − 10 = 0), or .

• f ( x ) = x − 3 and

The domain for f + g(x), fg (x), and fg(x) is [−2, ∞) (from x + 2 ≥ 0). The domain for is (−2, ∞) because we need .

## Function Composition

An important combination of two functions is function composition . This involves evaluating one function at the other. The notation for composing f with g is f o g(x). By definition, f o g(x) = f (g(x)), this means that we substitute g(x) for x in f(x).

#### Examples

Find f o g(x) and g o f(x).

• f (x) = x 2 + 1 and g (x) = 3 x + 2

f o g(x) = f(g(x))

= f (3 x + 2) Replace g (x) with 3 x + 2.

= (3 x + 2) 2 + 1 Substitute 3 x + 2 for x in f(x).

= (3x + 2)(3x + 2) + 1 = 9 x2 + 12 x + 5

g o f ( x ) = g (f (x))

= g(x2 + 1) Replace f(x) with x2 + 1.

= 3( x 2 + 1) + 2 Substitute x 2 + 1 for x in g(x).

= 3 x 2 + 3 + 2 = 3 x 2 + 5

•

At times, we only need to find f o g(x) for a particular value of x. The y-value for g(x) becomes the x -value for f(x).

#### Example

• Find f o g (−1), f o g (0), and g o f (1) for f ( x ) = 4 x + 3 and g ( x ) = 2−x2.
• f o g (−1) = f ( g (−1)) Compute g (−1).

= f (1) g (−1) = 2−(−1) 2 = 1

= 4(1) + 3 = 7 Evaluate f ( x ) at x = 1.

f o g (0) = f ( g (0)) Compute g (0).

= f (2) g (0) = 2 − 0 2 = 2

= 4(2) + 3 = 11 Evaluate f(x) at x = 2.

g o f (1) = g ( f (1)) Compute f (1).

= g (7) f (1) =4(1) + 3 = 7

= 2 − 7 2 = −47 Evaluate g ( x ) at x = 7.

We can compose two functions at a single x -value by looking at the graphs of the individual functions. To find f o g(a), we will look at the graph of g(x) to find the point whose x -coordinate is a. The y -coordinate of this point will be g(a). Then we will look at the graph of f(x) to find the point whose x -coordinate is g(a). The y -coordinate of this point will be f (g(a)) = f o g (a).

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