Function notation: evaluating and finding functions

A function is a rule that gives exactly one output for each allowed input. Function notation names the rule and makes the input explicit.

For example,

$f(x)=3x-4$

means: the function called $f$ multiplies its input by $3$, then subtracts $4$. If the input is $5$, the output is

$f(5)=3(5)-4=11.$

The notation $f(x)$ is read as “$f$ of $x$”. It does not mean $f\times x$.

Function notation occurs throughout A-level Mathematics, including graphs, transformations, calculus, sequences and numerical methods. The central habit is simple:

Replace every occurrence of the input variable by the complete new input.

Prerequisites

You should already be able to:

• substitute numbers into algebraic expressions;
• use negative numbers and the order of operations;
• expand brackets and collect like terms;
• solve simple linear and quadratic equations.

Review algebraic manipulation and factorisation if substitution or expansion is uncertain. Review linear equations or quadratic equations if solving the final equation causes difficulty.

Inputs, rules and outputs

Think of a function as a machine:

$$\text{input}\longmapsto\text{apply the rule}\longmapsto\text{output}.$$

If $f(x)=x^2+1$, then

$$2\longmapsto 2^2+1=5,$$

so $f(2)=5$. This can also be written as a mapping:

$f:2\mapsto5.$

The letter $f$ is merely the function’s name. Other names are equally valid:

$g(x)=2x+7,qquad h(t)=t^2-3t.$

In $h(t)$, the input variable is $t$. The choice of letter does not change the idea.

Function notation and equation notation

The statements

$y=3x-4$

and

$f(x)=3x-4$

can describe the same input and output rule. Function notation is more informative because it names the rule. It lets us write $f(5)$, $f(a)$ or $f(x+1)$ without inventing a new output variable each time.

Evaluating a function at a number

To evaluate $f(a)$, replace every $x$ in the formula for $f(x)$ by $a$, then simplify.

Worked example 1: a linear function

Given $f(x)=7-2x$, find $f(4)$.

Replace $x$ by $4$:

$$\begin{aligned} f(4)&=7-2(4)\\ &=7-8\\ &=-1. \end{aligned}$$

The expression $f(4)$ is the output, not an instruction to multiply $f$ by $4$.

Worked example 2: a negative input

Given $g(x)=2x^2-5x+1$, find $g(-3)$.

Use brackets around the negative input every time it replaces $x$:

$$\begin{aligned} g(-3)&=2(-3)^2-5(-3)+1\\ &=2(9)+15+1\\ &=34. \end{aligned}$$

Without brackets, $-3^2$ would mean $-(3^2)=-9$. The function requires the whole input $-3$ to be squared.

Worked example 3: a fractional input

Let

$h(x)=\frac{2x-1}{x+3}.$

Find $h\left(\frac12\right)$.

$$\begin{aligned} h\left(\frac12\right) &=\frac{2\left(\frac12\right)-1}{\frac12+3}\\ &=\frac{1-1}{\frac72}\\ &=0. \end{aligned}$$

The fraction bar groups the complete numerator and denominator. Substitute into both before simplifying.

Self-check 1

Let $p(x)=x^2-4x+6$. Find:

1. $p(3)$;
2. $p(-2)$;
3. $p(0)$.

Answers

1. $p(3)=3^2-4(3)+6=3$.
2. $p(-2)=(-2)^2-4(-2)+6=18$.
3. $p(0)=0^2-4(0)+6=6$.

The input can be an expression

In $f(3)$, the input is $3$. In $f(a)$, it is $a$. In $f(x+2)$, the complete input is $x+2$. The same replacement rule always applies.

Suppose

$f(x)=x^2-3x.$

Then

$f(\boxed{\phantom{x}})=(\boxed{\phantom{x}})^2-3(\boxed{\phantom{x}}).$

Putting $x+2$ into each box gives

$f(x+2)=(x+2)^2-3(x+2).$

The brackets protect the structure of the new input.

Worked example 4: substitute and expand

For $f(x)=x^2-3x$, simplify $f(x+2)$.

$$\begin{aligned} f(x+2) &=(x+2)^2-3(x+2)\\ &=x^2+4x+4-3x-6\\ &=x^2+x-2. \end{aligned}$$

A common incorrect first line is $x^2+2-3x+2$. This substitutes into separate pieces instead of replacing each whole occurrence of $x$. In particular,

$(x+2)^2\ne x^2+2.$

Worked example 5: a scaled input

Given $g(x)=4x^2+x-1$, simplify $g(2t)$.

$$\begin{aligned} g(2t) &=4(2t)^2+(2t)-1\\ &=4(4t^2)+2t-1\\ &=16t^2+2t-1. \end{aligned}$$

Because $(2t)^2=4t^2$, both the coefficient and variable are squared.

Worked example 6: compare nearby inputs

Let $f(x)=3x^2+2$. Find and simplify $f(x+1)-f(x)$.

First evaluate the two function expressions separately:

$$\begin{aligned} f(x+1)&=3(x+1)^2+2\\ &=3x^2+6x+5,\\[4pt] f(x)&=3x^2+2. \end{aligned}$$

Now subtract the complete second expression:

$$\begin{aligned} f(x+1)-f(x) &=(3x^2+6x+5)-(3x^2+2)\\ &=6x+3. \end{aligned}$$

This comparison of nearby outputs is an early version of an idea used later in differentiation.

Self-check 2

Given $q(x)=2x^2-5$, simplify:

1. $q(a)$;
2. $q(a+1)$;
3. $q(3a)$;
4. $q(-a)$.

Answers

1. $q(a)=2a^2-5$.
2. $q(a+1)=2(a+1)^2-5=2a^2+4a-3$.
3. $q(3a)=2(3a)^2-5=18a^2-5$.
4. $q(-a)=2(-a)^2-5=2a^2-5$.

The last answer equals $q(a)$ because squaring removes the sign. This does not happen for every function.

Finding an unknown function value

An equation such as $f(x)=17$ asks:

Which input or inputs produce the output $17$?

Replace $f(x)$ by its rule, then solve the resulting equation.

Worked example 7: solve a linear function equation

Given $f(x)=5x-3$, solve $f(x)=22$.

$$\begin{aligned} 5x-3&=22\\ 5x&=25\\ x&=5. \end{aligned}$$

Check:

$f(5)=5(5)-3=22.$

The answer is $x=5$, not $f=5$. The function is named $f$; the unknown is its input.

Worked example 8: there may be two inputs

Let $g(x)=x^2-4$. Solve $g(x)=12$.

$$\begin{aligned} x^2-4&=12\\ x^2&=16\\ x&=4\quad\text{or}\quad x=-4. \end{aligned}$$

Both inputs give the same output:

$g(4)=12,qquad g(-4)=12.$

A function assigns one output to each input. It may still assign the same output to several different inputs.

Worked example 9: an equation involving two function values

Given $h(x)=x^2+2x$, solve

$h(x)=h(3).$

First calculate the fixed value:

$h(3)=3^2+2(3)=15.$

Therefore

$$\begin{aligned} x^2+2x&=15\\ x^2+2x-15&=0\\ (x+5)(x-3)&=0. \end{aligned}$$

Hence

$x=-5\quad\text{or}\quad x=3.$

Do not cancel the letter $h$ from $h(x)=h(3)$. Function notation is not multiplication. Instead, evaluate both sides using the stated rule.

Finding the function rule

Sometimes the formula contains unknown constants. Given enough input and output information, form equations and solve for them.

Worked example 10: determine a linear rule

A function has the form

$f(x)=ax+b.$

Given $f(2)=7$ and $f(5)=16$, find $f(x)$.

Substitute each input and output:

$$\begin{aligned} 2a+b&=7,\\ 5a+b&=16. \end{aligned}$$

Subtract the first equation from the second:

$3a=9,$

so $a=3$. Use $2a+b=7$:

$2(3)+b=7,$

giving $b=1$. Therefore

$\boxed{f(x)=3x+1}.$

Check both facts:

$f(2)=7,qquad f(5)=16.$

Two points determine a linear function because its two constants, gradient and intercept, are then fixed.

Worked example 11: use a stated family of functions

The function $g$ has the form

$g(x)=x^2+px+q.$

Given $g(1)=4$ and $g(-2)=10$, find $p$ and $q$.

From $g(1)=4$,

$1+p+q=4,$

so

p+q=3.\tag{1}

From $g(-2)=10$,

$4-2p+q=10,$

so

-2p+q=6.\tag{2}

Subtract equation $(1)$ from equation $(2)$:

$-3p=3,$

giving $p=-1$. Then $p+q=3$ gives $q=4$. Thus

$\boxed{g(x)=x^2-x+4}.$

Tables and function notation

A function may be given by a table rather than a formula.

$x$	$-2$	$0$	$1$	$4$
$f(x)$	$5$	$1$	$-3$	$5$

From the table:

$f(0)=1,qquad f(4)=5.$

The equation $f(x)=5$ has two listed solutions:

$x=-2\quad\text{or}\quad x=4.$

The table gives no value for $f(3)$. Unless a rule or further information is supplied, it cannot be inferred. A function need not follow an obvious pattern merely because a few values are displayed.

Allowed inputs

Not every formula accepts every real number. The set of allowed inputs is called the domain.

For

$r(x)=\frac{1}{x-2},$

the input $x=2$ would make the denominator zero. Therefore $r(2)$ is undefined.

For

$s(x)=\sqrt{x+3},$

real outputs require $x+3\ge0$, so $x\ge-3$.

At this stage, develop the habit of checking for division by zero and square roots of negative numbers. Domains and ranges are treated fully in functions, domains and ranges.

Worked example 12: decide whether a value exists

Let

$k(x)=\frac{x+1}{x^2-9}.$

Determine whether $k(2)$ and $k(3)$ exist.

For $x=2$,

$k(2)=\frac{2+1}{2^2-9}=\frac3{-5}=-\frac35.$

For $x=3$, the denominator is

$3^2-9=0.$

Division by zero is undefined, so $k(3)$ does not exist.

Common misconceptions

• Reading $f(x)$ as $f\times x$. The letter $f$ names a rule. You cannot cancel or divide by it as though it were a factor.
• Replacing only some occurrences of the variable. In $f(x)=x^2+x$, both occurrences change when evaluating $f(a)$.
• Losing brackets around a negative or algebraic input. If $f(x)=x^2$, then $f(-3)=(-3)^2$, and $f(x+1)=(x+1)^2$.
• Confusing $f(x+1)$ with $f(x)+1$. If $f(x)=x^2$, then $f(x+1)=(x+1)^2$, while $f(x)+1=x^2+1$. They are generally different.
• Assuming one output has one input. A function requires one output per input, not one input per output. For $f(x)=x^2$, both $2$ and $-2$ produce $4$.
• Ignoring forbidden inputs. A formula containing a denominator or square root may have restrictions.

Mixed self-check

Let

$f(x)=2x^2-3x+1,qquad g(x)=4-x.$

1. Find $f(-2)$.
2. Simplify $g(3t-1)$.
3. Simplify $f(x+1)$.
4. Solve $g(x)=-5$.
5. Solve $f(x)=1$.
6. Find $a$ if $f(a)=f(2)$.

Answers

1.

$f(-2)=2(-2)^2-3(-2)+1=8+6+1=15.$

2.

$g(3t-1)=4-(3t-1)=5-3t.$

The subtraction changes both terms inside the bracket.

3.

$$\begin{aligned} f(x+1) &=2(x+1)^2-3(x+1)+1\\ &=2x^2+4x+2-3x-3+1\\ &=2x^2+x. \end{aligned}$$

4.

$4-x=-5,$

so $x=9$.

5.

$$\begin{aligned} 2x^2-3x+1&=1\\ 2x^2-3x&=0\\ x(2x-3)&=0. \end{aligned}$$

Therefore $x=0$ or $x=\frac32$.

6. First, $f(2)=8-6+1=3$. Now solve

$$\begin{aligned} 2a^2-3a+1&=3\\ 2a^2-3a-2&=0\\ (2a+1)(a-2)&=0. \end{aligned}$$

Hence

$a=-\frac12\quad\text{or}\quad a=2.$

What to learn next

You should now be able to explain $f(x)$ as an output, evaluate a function at numerical and algebraic inputs, solve equations involving function values, and determine simple function rules.

Next, study functions, domains and ranges for the full A-level language of mappings, domains, codomains, ranges and one-to-one functions. Then continue to composite and inverse functions and function graphs. If the algebraic manipulation was the main difficulty, strengthen algebraic fluency first.