Composition of Functions

In addition to adding, subtracting, multplying and dividing, two functions can be composed. The composition of a function is when the x-value is replaced by a function. For example if p(x) = x³ and q(x) = x - 1, the compostition of p with q is:

$p \circ q = p (q (x)) = p (x - 1) = {(x - 1)}^{3}$

The notation

p \circ q

, reads "p composed with q". Which means that the value of x is replaced by q(x) in function p.

THE DEFINITION OF COMPOSITION OF FUNCTIONS

The composition of the function p is

$(p \circ q) (x) = p (q (x))$

where the domain of $p \circ q$ is the set of all x in the domain of q such that q(x) is in the domain of p.

Let's look at a couple examples.

Example 1: Given p(x) = x² - 1 and q(x) = x + 3 find the composite function and its domain of $(p \circ q) (x)$ and $(q \circ p) (x)$ . Then evaluate $(p \circ q) (3)$ .

Step 1: Find $(p \circ q) (x)$ and its domain.

$(p \circ q) (x) = p (q (x))$ Def'n of compos.

$p (x + 3) = {(x + 3)}^{2} - 1$ Sub. q(x)

$(p \circ q) = x^{2} + 6 x + 9 - 1$ Simplify

$(p \circ q) = x^{2} + 6 x + 8$

The domain of p(x) = x² - 1 is the set of all real numbers.

The domain of q(x) = x + 3 is the set of all real numbers.

Therefore the domain of $(p \circ q)$ is the set of all real numbers.

Step 2: Find $(q \circ p) (x)$ and its domain.

$(q \circ p) (x) = q (p (x))$ Def'n of compos.

$q (x^{2} - 1) = (x^{2} - 1) + 3$ Sub. p(x)

$(q \circ p) = x^{2} + 2$ Simplify

The domain of q(x) = x + 3 is the set of all real numbers.

The domain of p(x) = x² - 1 is the set of all real numbers.

Therefore the domain of $(q \circ p)$ is the set of all real numbers.

Step 3: Evaluate $(p \circ q) (x)$ when x = 3.

$(p \circ q) (x) = x^{2} + 6 x + 8$

$(p \circ q) (3) = 3^{2} + 6 (3) + 8 = 35$

Example 2: Given $p (x) = \sqrt{4 x}$ and q(x) = x² find the composite function and its domain of $(p \circ q) (x)$ and $(q \circ p) (x)$ . Then evaluate $(p \circ q) (- 5)$ .

Step 1: Find $(p \circ q) (x)$ and its domain.

$(p \circ q) (x) = p (q (x))$ Def'n of compos.

$p (x^{2}) = \sqrt{4 x^{2}}$ Sub. q(x)

$(p \circ q) = 2 x$ Simplify

The domain of $p (x) = \sqrt{4 x}$ is the set of all real numbers such that $x \geq 0$ .

The domain of q(x) = x² is the set of all real numbers.

Therefore the domain of $(p \circ q)$ is the set of all real numbers such that $x \geq 0$ .

Step 2: Find $(q \circ p) (x)$ and its domain.

$(q \circ p) (x) = q (p (x))$ Def'n of compos.

$q (\sqrt{4 x}) = {(\sqrt{4 x})}^{2}$ Sub. p(x)

$(q \circ p) = 4 x$ Simplify

The domain of q(x) = x² is the set of all real numbers.

The domain of $p (x) = \sqrt{4 x}$ is the set of all real numbers such that $x \geq 0$ .

Therefore the domain of $(p \circ q)$ is the set of all real numbers such that $x \geq 0$ .

Step 3: Evaluate $(q \circ p) (x)$ when x = -5.

$(p \circ q) (x) = 4 x$

$(p \circ q) (- 5) = 4 (- 5) = - 20$

Related Links:
Math
algebra
Extreme Value Theorem
Even and Odd Functions
Calculus Topics

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