Define the decomposition of a function.

The decomposition of a function is a way of rewriting it as the composition of two or more simpler functions. Instead of one complicated rule you end up with an inner function and an outer function that together produce the same outputs.

What is the horizontal line test, and what does it tell you?

Check whether any horizontal line crosses the graph more than once. If one does, then two different inputs share an output, so the function is not invertible on that domain.

True or False? A function that is always increasing on its domain is automatically invertible there.

True. If the output values only ever move in one direction, no two different inputs can share an output. The same argument works for a function that is always decreasing.

AP®MathsCollege BoardPrecalculusFlashcardsExponential & Logarithmic FunctionsComposite & Inverse Functions

Composite & Inverse Functions (College Board AP® Precalculus): Flashcards

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FrontComposition of Functions
Define composite function.

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Cards in this collection (44)

Define composite function.
A composite function $f \circ g$ uses the output values of $g$ as the input values for $f$ .
It is also written $f (g (x))$ , and read aloud as $f$ of $g$ of $x$ .
Fill in the two gaps.
In the composite function $f (g (x))$ , the function $g$ is called the $______$ function and $f$ is called the $______$ function.
The completed sentence is:
In the composite function $f (g (x))$ , the function $g$ is called the inner function and $f$ is called the outer function.
True or False?
In $(f \circ g) (x)$ , the function $f$ acts on the input first.
False.
The function on the right acts on the input first, so $g$ takes the input and its output is then fed into $f$ .
That is why $(f \circ g) (x)$ is also written $f (g (x))$ , with $g$ innermost.
The functions are $f (x) = 2 x + 3$ and $g (x) = x^{2} - 1$ . Find $f (g (2))$ .
$f (g (2)) = 9$ .
First work out $g (2) = 2^{2} - 1 = 3$ , then feed that in to get $f (3) = 2 (3) + 3 = 9$ .
What is the domain of the composite function $f \circ g$ ?
It is restricted to those input values of $g$ whose output values lie in the domain of $f$ .
If $g (x)$ produces a value that is not in the domain of $f$ , then $f (g (x))$ is not defined at that input.
How do you evaluate a composite function when $g$ is given by a table and $f$ by a formula?
Read the value of $g (x)$ from the table, then substitute that value into the formula for $f$ .
For instance, if the table gives $g (2) = 5$ , the answer is whatever the formula gives for $f (5)$ .
Is the composition of functions commutative?
No, $f \circ g$ and $g \circ f$ are generally different functions.
With $f (x) = x^{2}$ and $g (x) = x + 3$ , $f (g (1)) = 16$ but $g (f (1)) = 4$ .
Define the identity function.
The identity function is the function given by $f (x) = x$ .
It maps every input to itself, so the output is always equal to the input.
What happens when the identity function is composed with another function?
The other function is left completely unchanged, in either order.
So the identity function behaves in composition the way $0$ behaves in addition and $1$ behaves in multiplication.
Fill in the two gaps.
The function $T$ gives temperature as a function of time, and $P$ gives pressure as a function of temperature. Pressure as a function of time is found by applying $______$ to the output of $______$ at each time.
The completed sentence is:
Pressure as a function of time is found by applying $P$ to the output of $T$ at each time.
Composing two functions this way relates two quantities that no single existing formula connects directly.
How do you construct $f (g (x))$ from the formulas for $f$ and $g$ ?
Write out the formula for $f (x)$ , replace every $x$ in it with the entire expression for $g (x)$ , then simplify.
With $f (x) = 3 x + 1$ and $g (x) = x^{2} - 4$ this gives $3 (x^{2} - 4) + 1 = 3 x^{2} - 11$ .
True or False?
To construct $f (g (x))$ it is enough to replace one occurrence of $x$ in the formula for $f$ .
False.
Every occurrence of $x$ in $f$ has to be replaced, which is the commonest error in this topic.
For $f (x) = x^{2} + 2 x$ and $g (x) = x - 3$ , both terms need it: ${(x - 3)}^{2} + 2 (x - 3) = x^{2} - 4 x + 3$ .
The functions are $f (x) = x^{2} - 2$ and $g (x) = \frac{x + 1}{x}$ . Find an expression for $f (g (x))$ .
$f (g (x)) = \frac{1 + 2 x - x^{2}}{x^{2}}$ .
Substituting gives ${(\frac{x + 1}{x})}^{2} - 2 = \frac{x^{2} + 2 x + 1}{x^{2}} - 2$ .
Writing $2$ as $\frac{2 x^{2}}{x^{2}}$ and combining the two fractions leaves $\frac{1 + 2 x - x^{2}}{x^{2}}$ .
How do you build a table of values for $f \circ g$ ?
For each input $x$ , work out $g (x)$ first, then find $f$ at that value.
Record the resulting pairs $(x, f (g (x)))$ to build up the table.
How do you build a graph of $f \circ g$ when $f$ and $g$ are themselves given as graphs?
For a selection of input values, read $g (x)$ from the graph of $g$ , then read $f$ at that value from the graph of $f$ .
Plot the points you get and sketch a curve through them.
Define the decomposition of a function.
The decomposition of a function is a way of rewriting it as the composition of two or more simpler functions.
Instead of one complicated rule you end up with an inner function and an outer function that together produce the same outputs.
How do you find a decomposition of a function given by a formula?
Pick an inner part of the expression and call it $g (x)$ , then replace every instance of that part with a plain $x$ .
Whatever is left is the outer function $f (x)$ .
For $h (x) = \sqrt{x^{2} - 5}$ , taking $g (x) = x^{2} - 5$ leaves $f (x) = \sqrt{x}$ .
The function $h$ is given by $h (x) = {(2 x + 1)}^{3}$ . Fill in the two gaps to decompose it as $h (x) = f (g (x))$ , where the first gap is an exponent.
$f (x) = x^{______} and g (x) =______$
The completed decomposition is:
$f (x) = x^{3} and g (x) = 2 x + 1$
The inner function is the part inside the bracket, and the outer function is the cubing that is applied to it.
How do you check that a decomposition is correct?
Substitute the inner function back into the outer one, replacing every $x$ in $f$ with the whole of $g (x)$ .
If the result matches the original function, the decomposition is right.
True or False?
A function has only one correct decomposition.
False.
Decompositions are not unique, and the same function can often be broken up in more than one way.
For $h (x) = {(x + 3)}^{2} + 5$ you could take $f (x) = x^{2} + 5$ with $g (x) = x + 3$ , or $f (x) = x + 5$ with $g (x) = {(x + 3)}^{2}$ .
Composing $f$ with $g (x) = x + k$ gives a translation. Why does the order of composition decide whether it is horizontal or vertical?
Because the order decides whether the constant reaches the input or the output.
$f (g (x)) = f (x + k)$ adds $k$ to the input, while $g (f (x)) = f (x) + k$ adds it to the output instead.
The function $h$ is given by $h (x) = 3 {(x + 4)}^{2} - 7$ . Write it as $p (q (x))$ with $q$ linear, and relate the graph of $h$ to the graph of $p$ .
Take $q (x) = x + 4$ and $p (x) = 3 x^{2} - 7$ .
Since $p (q (x)) = p (x + 4)$ , the constant reaches the input, so the graph of $h$ is the graph of $p$ translated $4$ units to the left.
Define inverse function.
An inverse function reverses the mapping of the original function.
If $f (a) = b$ then $f^{- 1} (b) = a$ , so an input-output pair $(a, b)$ of $f$ becomes the pair $(b, a)$ of $f^{- 1}$ .
True or False?
The notation $f^{- 1} (x)$ means $\frac{1}{f (x)}$ .
False.
The notation $f^{- 1}$ means the inverse function of $f$ , and not a reciprocal.
The superscript $- 1$ is part of the name of the function here rather than an exponent.
Fill in the two gaps.
A function is invertible on a given domain when every $______$ value is mapped from a unique $______$ value.
The completed sentence is:
A function is invertible on a given domain when every output value is mapped from a unique input value.
No two different inputs may produce the same output, because the reverse mapping has itself to be a function.
What is the horizontal line test, and what does it tell you?
Check whether any horizontal line crosses the graph more than once.
If one does, then two different inputs share an output, so the function is not invertible on that domain.
A table shows that $g (2) = 3$ . What is $g^{- 1} (3)$ , and how did you get it?
$g^{- 1} (3) = 2$ .
Read the table backwards: find $3$ in the output row, then take the input value it was produced from.
The function $f (x) = x^{2}$ is not invertible on all real numbers. How can it be made invertible?
Restrict the domain to an interval on which no output value is repeated, for example $x \geq 0$ .
On that restricted domain each output comes from a single input, and $f^{- 1} (x) = \sqrt{x}$ .
There is usually more than one way to choose such a restriction.
True or False?
A function that is always increasing on its domain is automatically invertible there.
True.
If the output values only ever move in one direction, no two different inputs can share an output.
The same argument works for a function that is always decreasing.
A table for $f$ shows $f (- 4) = 10$ and $f (4) = 10$ . How do you show that $f$ is not invertible?
Name the repeated output explicitly: the value $10$ is produced by both $- 4$ and $4$ .
Because that output does not come from a unique input, the reverse mapping would not be a function, so $f$ has no inverse on this domain.
Fill in the two gaps.
When $f$ is invertible, the domain of $f$ is the $______$ of $f^{- 1}$ , and the range of $f$ is the $______$ of $f^{- 1}$ .
The completed sentence is:
When $f$ is invertible, the domain of $f$ is the range of $f^{- 1}$ , and the range of $f$ is the domain of $f^{- 1}$ .
Inputs and outputs swap roles, so the two sets simply change places.
The function $f (x) = \frac{4 x + 7}{2}$ has domain $[- 3, 5]$ . Find the domain of $f^{- 1}$ .
The domain of $f^{- 1}$ is $[- 2.5, 13.5]$ .
You need the range of $f$ first, and since $f$ is an increasing linear function only the endpoints matter: $f (- 3) = - 2.5$ and $f (5) = 13.5$ .
That range is then the domain of the inverse.
True or False?
Once $f$ is known to be invertible, $f^{- 1}$ is meaningful for every value in its domain.
False.
In a real-world context, further restrictions can limit where the inverse is actually useful.
If $f (t)$ gives the height of a ball at time $t$ , then $f^{- 1} (h)$ means something only for heights the ball really reaches, and only while it is in the air.
How do you build a table of values for $f^{- 1}$ from a table of values for $f$ ?
Reverse the roles of the two rows, so the inputs of $f$ become the outputs of $f^{- 1}$ and the outputs of $f$ become its inputs.
Every value already in the table for $f$ therefore appears in the table for $f^{- 1}$ , just in the other row.
Fill in the two gaps.
To find an inverse from a formula, start by writing $y = f (x)$ , then $______$ the two variables, and finally $______$ the result for $y$ .
The completed sentence is:
To find an inverse from a formula, start by writing $y = f (x)$ , then swap the two variables, and finally solve the result for $y$ .
What you end up with is $y = f^{- 1} (x)$ , and the method works by reversing the operations $f$ performs on its input.
Find the inverse of $f (x) = 2 x^{3} - 5$ .
$f^{- 1} (x) = \sqrt[3]{\frac{x + 5}{2}}$ .
Write $y = 2 x^{3} - 5$ , swap the variables to get $x = 2 y^{3} - 5$ , then rearrange to $y^{3} = \frac{x + 5}{2}$ .
Taking the cube root of both sides gives the answer.
What do you get when you compose a function with its inverse, and why?
You get back the input you started with, so $f (f^{- 1} (x)) = f^{- 1} (f (x)) = x$ .
That happens because $f^{- 1}$ reverses exactly the mapping $f$ performs, so the two undo each other.
The result is the identity function, in either order.
You believe $f^{- 1} (x) = \frac{x + 5}{3}$ is the inverse of $f (x) = 3 x - 5$ . How can you check?
Compose the two functions and check that the result simplifies to $x$ .
Here $f (f^{- 1} (x)) = 3 \cdot \frac{x + 5}{3} - 5 = (x + 5) - 5 = x$ , which confirms it.
How is the graph of $y = f^{- 1} (x)$ related to the graph of $y = f (x)$ ?
It is the reflection of the graph of $f$ over the line $y = x$ .
That line is the graph of the identity function, which is why reflecting in it undoes what $f$ does.
Fill in the two gaps.
If the point $(a, b)$ lies on the graph of $f$ , then the point $(______,______)$ lies on the graph of $f^{- 1}$ .
The completed statement is:
If the point $(a, b)$ lies on the graph of $f$ , then the point $(b, a)$ lies on the graph of $f^{- 1}$ .
The two coordinates change places, so a point high up on the graph of $f$ sits far to the right on the graph of $f^{- 1}$ .
What happens to the intercepts and the asymptotes of $f$ on the graph of $f^{- 1}$ ?
They change direction: $x$ -intercepts of $f$ become $y$ -intercepts of $f^{- 1}$ , and $y$ -intercepts become $x$ -intercepts.
In the same way, vertical asymptotes of $f$ become horizontal asymptotes of $f^{- 1}$ , and horizontal ones become vertical.
True or False?
If $f$ is increasing, then $f^{- 1}$ is decreasing.
False.
If $f$ is increasing then $f^{- 1}$ is increasing as well.
Reflecting over $y = x$ does not turn a rising graph into a falling one, because larger outputs still came from larger inputs.
How do you read the value of $f^{- 1} (k)$ from the graph of $f$ ?
Find $k$ on the $y$ -axis, read across to the curve, then read down to the $x$ -axis.
The common mistake is to look $k$ up on the $x$ -axis instead, which gives $f (k)$ rather than $f^{- 1} (k)$ .
The graph of $f$ is increasing and passes through $(0, 0)$ , $(2, 2)$ , $(3, 6)$ and $(4, 8)$ . What is the maximum value of $f^{- 1}$ ?
The maximum value of $f^{- 1}$ is $4$ .
Reflecting the highest point $(4, 8)$ over $y = x$ gives $(8, 4)$ , which is the highest point on the graph of $f^{- 1}$ .
The outputs of the inverse are the inputs of $f$ , so they stop at $4$ .

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