Because we cannot take the square root of a negative number, the domain of
$\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ is
$\text{\hspace{0.17em}}\left(-\infty ,3\right].\text{\hspace{0.17em}}$ Now we check the domain of the composite function
The domain of this function is
$\text{\hspace{0.17em}}\left(-\infty ,5\right].\text{\hspace{0.17em}}$ To find the domain of
$\text{\hspace{0.17em}}f\circ g,\text{\hspace{0.17em}}$ we ask ourselves if there are any further restrictions offered by the domain of the composite function. The answer is no, since
$\text{\hspace{0.17em}}\left(-\infty ,3\right]\text{\hspace{0.17em}}$ is a proper subset of the domain of
$\text{\hspace{0.17em}}f\circ g.\text{\hspace{0.17em}}$ This means the domain of
$\text{\hspace{0.17em}}f\circ g\text{\hspace{0.17em}}$ is the same as the domain of
$\text{\hspace{0.17em}}g,\text{\hspace{0.17em}}$ namely,
$\text{\hspace{0.17em}}\left(-\infty ,3\right].$
Decomposing a composite function into its component functions
In some cases, it is necessary to decompose a complicated function. In other words, we can write it as a composition of two simpler functions. There may be more than one way to
decompose a composite function , so we may choose the decomposition that appears to be most expedient.
Decomposing a function
Write
$\text{\hspace{0.17em}}f(x)=\sqrt{5-{x}^{2}}\text{\hspace{0.17em}}$ as the composition of two functions.
We are looking for two functions,
$\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ and
$\text{\hspace{0.17em}}h,\text{\hspace{0.17em}}$ so
$\text{\hspace{0.17em}}f(x)=g(h(x)).\text{\hspace{0.17em}}$ To do this, we look for a function inside a function in the formula for
$\text{\hspace{0.17em}}f(x).\text{\hspace{0.17em}}$ As one possibility, we might notice that the expression
$\text{\hspace{0.17em}}5-{x}^{2}\text{\hspace{0.17em}}$ is the inside of the square root. We could then decompose the function as
$$h(x)=5-{x}^{2}\text{and}g(x)=\sqrt{x}$$
We can check our answer by recomposing the functions.
We can perform algebraic operations on functions. See
[link] .
When functions are combined, the output of the first (inner) function becomes the input of the second (outer) function.
The function produced by combining two functions is a composite function. See
[link] and
[link] .
The order of function composition must be considered when interpreting the meaning of composite functions. See
[link] .
A composite function can be evaluated by evaluating the inner function using the given input value and then evaluating the outer function taking as its input the output of the inner function.
A composite function can be evaluated from a table. See
[link] .
A composite function can be evaluated from a graph. See
[link] .
A composite function can be evaluated from a formula. See
[link] .
The domain of a composite function consists of those inputs in the domain of the inner function that correspond to outputs of the inner function that are in the domain of the outer function. See
[link] and
[link] .
Just as functions can be combined to form a composite function, composite functions can be decomposed into simpler functions.
Functions can often be decomposed in more than one way. See
[link] .
Section exercises
Verbal
How does one find the domain of the quotient of two functions,
$\text{\hspace{0.17em}}\frac{f}{g}?\text{\hspace{0.17em}}$
Find the numbers that make the function in the denominator
$\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ equal to zero, and check for any other domain restrictions on
$\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ and
$\text{\hspace{0.17em}}g,\text{\hspace{0.17em}}$ such as an even-indexed root or zeros in the denominator.
not much
For functions, there are two conditions for a function to be the inverse function:
1--- g(f(x)) = x for all x in the domain of f
2---f(g(x)) = x for all x in the domain of g Notice in both cases you will get back to the element that you started with, namely, x.