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Operations on Functions

Functions with overlapping domains can be added, subtracted, multiplied and divided. If $f (x)$ and $g (x)$ are two functions, then for all $x$ in the domain of both functions the sum, difference, product and quotient are defined as follows.

$\begin{array}{l} (f + g) (x) = f (x) + g (x) \\ (f - g) (x) = f (x) - g (x) \\ (f g) (x) = f (x) \times g (x) \\ (\frac{f}{g}) (x) = \frac{f (x)}{g (x)}, g (x) \neq 0 \end{array}$

Example :

Let $f (x) = 2 x + 1$ and $g (x) = x^{2} - 4$

Find $(f + g) (x), (f - g) (x), (f g) (x)$ and $(\frac{f}{g}) (x)$ .

$\begin{array}{l} (f + g) (x) = f (x) + g (x) \\ = (2 x + 1) + (x^{2} - 4) \\ = x^{2} + 2 x - 3 \end{array}$

$\begin{array}{l} (f - g) (x) = f (x) - g (x) \\ = (2 x + 1) - (x^{2} - 4) \\ = - x^{2} + 2 x + 5 \end{array}$

$\begin{array}{l} (f g) (x) = f (x) \times g (x) \\ = (2 x + 1) (x^{2} - 4) \\ = 2 x^{3} + x^{2} - 8 x - 4 \end{array}$

$(\frac{f}{g}) (x) = \frac{f (x)}{g (x)} = \frac{2 x + 1}{x^{2} - 4}, x \neq \pm 2$

Another way to combine two functions to create a new function is called the composition of functions . In the composition of functions we substitute an entire function into another function.

The notation of the function $f$ with $g$ is $(f \circ g) (x) = f (g (x))$ and is read $f$ of $g$ of $x$ . It means that wherever there is an $x$ in the function $f$ , it is replaced with the function $g (x)$ . The domain of $f \circ g$ is the set of all $x$ in the domain of $g$ such that $g (x)$ is in the domain of $f$ .

Example 1:

Let $f (x) = x^{2}$ and $g (x) = x - 3$ . Find $f (g (x))$ .

$\begin{array}{l} f (g (x)) = f (x - 3) \\ = {(x - 3)}^{2} \\ = x^{2} - 6 x + 9 \end{array}$

Example 2:

Let $f (x) = 2 x - 1$ and $g (x) = x + 2$ . Find $f (g (x))$ .

$\begin{array}{l} f (g (x)) = f (x + 2) \\ = 2 (x + 2) - 1 \\ = 2 x + 3 \end{array}$

Order DOES matter when finding the composition of functions.

Example 3:

Let $f (x) = 3 x + 1$ and $g (x) = 2 x - 3$ .

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

$\begin{array}{l} f (g (x)) = f (2 x - 3) \\ = 3 (2 x - 3) + 1 \\ = 6 x - 8 \\ g (f (x)) = f (3 x + 1) \\ = 2 (3 x + 1) - 3 \\ = 6 x - 1 \end{array}$

Since $6 x - 8 \neq 2 x - 1, f (g (x)) \neq g (f (x))$ .