Define a piecewise function.

A piecewise function is a single function made of different parts, each with its own domain. To evaluate it, first decide which domain the input lies in, then use that part's rule.

True or False? A piecewise function's graph can have a jump in it.

True. The parts do not have to join up, and a jump like that is called a discontinuity . Plotting the end-points of each part is what shows where any jumps happen.

What does an inverse function do?

An inverse function undoes the original one, applying the inverse operations in the reverse order. So the inverse of "double and add 1" is "subtract 1 and halve".

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Functions (AQA GCSE Further Maths): Flashcards

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In $f (x) = 2 x + 1$ , what do $x$ and $f (x)$ stand for?

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In $f (x) = 2 x + 1$ , what do $x$ and $f (x)$ stand for?
$x$ is the input and $f (x)$ is the output.
Whatever goes inside the bracket replaces every $x$ on the other side, so $f (3) = 2 \times 3 + 1 = 7$ .
If $f (x) = 3 x^{2} - 2 x + 1$ , find $f (x + 3)$ .
Replace every $x$ with $(x + 3)$ , giving $3 {(x + 3)}^{2} - 2 (x + 3) + 1$ .
Expanding and simplifying gives $3 x^{2} + 16 x + 22$ .
If $g (x) = 3 x - 4$ , find $x$ when $g (x) = - 16$ .
Set the function equal to the output and solve, so $3 x - 4 = - 16$ .
That gives $3 x = - 12$ , so $x = - 4$ .
Define the domain and the range of a function.
The domain is the set of all inputs the function is allowed to take, described in terms of $x$ .
The range is the set of all outputs it gives out, described in terms of $f (x)$ .
True or False?
A range must be written in terms of $f (x)$ rather than $y$ .
True.
A range describes the outputs, which the function notation names directly.
You may sketch with $y$ in order to see the range, but rewrite it as $f (x)$ in the answer.
What is the domain of $f (x) = \sqrt{x - 2}$ , and why?
The domain is $x \geq 2$ , because you cannot square-root a negative number.
It comes from solving $x - 2 \geq 0$ .
What is the domain of $f (x) = \frac{1}{x - 5}$ ?
The domain is all real values of $x$ except 5, because dividing by zero is not allowed.
A domain can exclude single values like this, using the word "except".
For $g (x) = 9 - x$ the range is $4 < g (x) \leq 6$ . Why is the domain $3 \leq x < 5$ rather than $3 < x \leq 5$ ?
Because the function decreases, so the largest output comes from the smallest input.
The output 6 comes from $x = 3$ and carries its $\leq$ to that end, while the output just above 4 comes from just below 5.
Define a piecewise function.
A piecewise function is a single function made of different parts, each with its own domain.
To evaluate it, first decide which domain the input lies in, then use that part's rule.
$f (x)$ is $x^{2}$ on $- 2 \leq x \leq 2$ , 4 on $2 < x \leq 5$ , and $9 - x$ on $5 < x \leq 9$ . Find $f (8.5)$ .
8.5 lies in $5 < x \leq 9$ , so the third rule is the one that applies.
That gives $f (8.5) = 9 - 8.5 = 0.5$ .
True or False?
A piecewise function's graph can have a jump in it.
True.
The parts do not have to join up, and a jump like that is called a discontinuity.
Plotting the end-points of each part is what shows where any jumps happen.
In $fg (x)$ , which function is applied first?
$g$ is applied first, because it is the letter nearest the bracket.
$fg (x)$ means $f (g (x))$ , so f acts on the output of g.
If $f (x) = 2 x - 1$ and $g (x) = {(x + 2)}^{2}$ , find $fg (4)$ .
Apply g first, so $g (4) = {(4 + 2)}^{2} = 36$ .
Then $f (36) = 2 \times 36 - 1 = 71$ .
If $f (x) = 2 x - 1$ and $g (x) = {(x + 2)}^{2}$ , find $gf (x)$ .
Substitute $f (x)$ into g, which gives ${((2 x - 1) + 2)}^{2}$ .
Simplifying inside the bracket gives ${(2 x + 1)}^{2}$ .
True or False?
$fg (x)$ and $gf (x)$ always give the same result.
False.
The order matters, so they are usually different functions altogether.
With $f (x) = 2 x + 1$ and $g (x) = \frac{1}{x}$ , $fg (2) = 2$ but $gf (2) = \frac{1}{5}$ .
What does an inverse function do?
An inverse function undoes the original one, applying the inverse operations in the reverse order.
So the inverse of "double and add 1" is "subtract 1 and halve".
How do you find the inverse of $f (x) = 5 - 3 x$ ?
Write $y = 5 - 3 x$ , swap the $x$ and $y$ to get $x = 5 - 3 y$ , then rearrange for $y$ .
That gives $f^{- 1} (x) = \frac{5 - x}{3}$ .
Complete the property of a function and its inverse:
${ff}^{- 1} (x) = f^{- 1} f (x) =______$
The completed property is:
${ff}^{- 1} (x) = f^{- 1} f (x) = x$
Applying a function and then its inverse undoes whatever happened, so you end up back where you started.
If $f (3) = 10$ , what is $f^{- 1} (10)$ ?
$f^{- 1} (10) = 3$ , because the inverse swaps the input and the output.
An input of 3 gave an output of 10, so an input of 10 into the inverse gives 3 back.
If $f (x) = 2^{x}$ , how can you solve $f^{- 1} (x) = 5$ without finding the inverse?
Apply f to both sides, since ${ff}^{- 1}$ cancels, leaving $x = f (5)$ .
That gives $x = 2^{5} = 32$ .

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