Function Notation

Function notation, like $f(x)$, gives a function a name and shows its input. It’s a short way to say “the output of $f$ when the input is $x$”, and it makes it easy to work with several functions at once.

Key ideas

Reading f(x)

$f(x)$ is read ”$f$ of $x$”.

• $f$ is the name of the function. Other letters are fine too: $g$, $h$, $C$ for cost, $h$ for height.
• $x$ is the input.
• $f(x)$ is the output.

$f(x)$ does not mean $f$ times $x$.

Since the output is the $y$-value, $y = f(x)$, and each point on the graph has the form $(x, f(x))$.

Evaluating a function

To evaluate, replace every $x$ with the input, in brackets, then simplify. If $f(x) = x^2 - 3x$:

$$f(-2) = (-2)^2 - 3(-2) = 4 + 6 = 10$$

So the point $(-2, 10)$ is on the graph of $y = f(x)$.

You can also substitute an expression. For $f(a + 1)$, replace every $x$ with $(a + 1)$.

Two different questions

• Find $f(2)$: you know the input ($2$) and want the output.
• Solve $f(x) = 2$: you know the output ($2$) and want the input(s). There may be more than one answer.

Reading a graph

• $f(3)$ is the height of the graph at $x = 3$.
• To solve $f(x) = 3$, find every point on the graph at height $3$ and read their $x$-values.

On the SAT

When the SAT gives a function as an equation, Desmos evaluates it fast: type f(x) = 3x^2 - 5x + 1 on one line and f(4) on the next, and the value appears; you can even type f(4) - f(-2). To solve $f(x) = 10$, also graph y = 10 and click the intersections; the answers are their x-coordinates. When the function is given as a table or a graph, just read it: that’s faster than entering it into Desmos. See using Desmos on the SAT.

Worked examples

Example 1: Evaluating

Let $f(x) = 3x - 5$ and $g(x) = x^2 - 2x$. Find $f(4)$, $g(-3)$, and $f(0) + g(1)$.

Solution.

$$f(4) = 3(4) - 5 = 7$$ $$g(-3) = (-3)^2 - 2(-3) = 9 + 6 = 15$$ $$f(0) + g(1) = \big(3(0) - 5\big) + \big(1^2 - 2(1)\big) = -5 + (-1) = -6$$

Example 2: Substituting expressions

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

Solution. Replace every $x$ with the new input, in brackets.

$$\begin{aligned} f(a + 1) &= 2(a + 1)^2 - (a + 1) + 1 \\ &= 2(a^2 + 2a + 1) - a - 1 + 1 \\ &= 2a^2 + 4a + 2 - a \\ &= 2a^2 + 3a + 2 \end{aligned}$$ $$f(2x) = 2(2x)^2 - (2x) + 1 = 8x^2 - 2x + 1$$

Example 3: Solving f(x) = k

Using $f$ and $g$ from Example 1, solve $f(x) = 10$ and $g(x) = 8$.

Solution.

$$3x - 5 = 10 \quad\Rightarrow\quad 3x = 15 \quad\Rightarrow\quad x = 5$$ $$\begin{aligned} x^2 - 2x &= 8 \\ x^2 - 2x - 8 &= 0 \\ (x - 4)(x + 2) &= 0 \end{aligned}$$

So $x = 4$ or $x = -2$. Both inputs give an output of $8$: check $g(4) = 16 - 8 = 8$ and $g(-2) = 4 + 4 = 8$.

Example 4: Reading a graph

Use the graph of $y = h(x)$ below to find $h(3)$ and $h(1)$, and to solve $h(x) = 3$.

Solution.

• At $x = 3$ the graph is at height $0$, so $h(3) = 0$.
• At $x = 1$ the graph is at its highest point, height $4$, so $h(1) = 4$.
• The dashed line $y = 3$ crosses the graph at $(0, 3)$ and $(2, 3)$, so $h(x) = 3$ when $x = 0$ or $x = 2$.

Common mistakes

Treating $f(x)$ as multiplication. $f(3)$ means “the output when the input is $3$”, not $f \times 3$.

Dropping brackets with negative numbers. For $f(x) = x^2$, $f(-3) = (-3)^2 = 9$. Without brackets you’d get $-3^2 = -9$, which is wrong.

Mixing up $f(2)$ and $f(x) = 2$. The first gives you an output. The second asks you to find input(s).

Substituting an expression without brackets. For $f(x) = 2x^2$, $f(a + 1) = 2(a + 1)^2$, not $2a + 1^2$.

Stopping at one answer. Equations like $x^2 - 2x = 8$ can have two solutions. Check whether there’s another.

Practice

1. (Warm-up) Let $f(x) = 5 - 2x$. Find $f(3)$, $f(-1)$, and $f(0)$.

Solution

$$f(3) = 5 - 6 = -1, \qquad f(-1) = 5 + 2 = 7, \qquad f(0) = 5$$

2. (Warm-up) Let $g(x) = x^2 + 4x$. Find $g(-2)$ and $g\!\left(\dfrac{1}{2}\right)$.

Solution

$$g(-2) = (-2)^2 + 4(-2) = 4 - 8 = -4$$$$g\!\left(\tfrac{1}{2}\right) = \tfrac{1}{4} + 2 = \tfrac{9}{4}$$

3. (Warm-up) Use the table to find $f(2)$ and to solve $f(x) = 13$.

$x$	$0$	$1$	$2$	$3$
$f(x)$	$4$	$7$	$10$	$13$

Solution

$f(2) = 10$. The output $13$ appears when $x = 3$, so $f(x) = 13$ when $x = 3$.

4. (Core) Let $f(x) = 3x + 2$. Solve $f(x) = -10$.

Solution

$$3x + 2 = -10 \quad\Rightarrow\quad 3x = -12 \quad\Rightarrow\quad x = -4$$

5. (Core) Let $h(x) = x^2 - 6x$. Solve $h(x) = -5$.

Solution

$$x^2 - 6x + 5 = 0 \quad\Rightarrow\quad (x - 1)(x - 5) = 0$$

So $x = 1$ or $x = 5$. Check: $h(1) = 1 - 6 = -5$ and $h(5) = 25 - 30 = -5$.

6. (Core) Let $f(x) = x^2 - 3x$. Find and simplify $f(a - 2)$.

Solution

$$\begin{aligned} f(a - 2) &= (a - 2)^2 - 3(a - 2) \\ &= a^2 - 4a + 4 - 3a + 6 \\ &= a^2 - 7a + 10 \end{aligned}$$

7. (Core) A school club orders T-shirts. The cost in dollars for $n$ shirts is $C(n) = 8n + 45$.

• (a) Find $C(20)$ and explain what it means.
• (b) Solve $C(n) = 285$ and explain what it means.

Solution

(a) $C(20) = 8(20) + 45 = 205$. Ordering 20 shirts costs $205.

(b) $8n + 45 = 285$, so $8n = 240$ and $n = 30$. For $285, the club can order 30 shirts.

8. (Challenge) Let $f(x) = ax + 3$. If $f(2) = 11$, find $a$, then find $f(-1)$.

Solution

$f(2) = 2a + 3 = 11$, so $2a = 8$ and $a = 4$.

Then $f(x) = 4x + 3$, so $f(-1) = -4 + 3 = -1$.

9. (Challenge) Let $f(x) = 2x - 1$ and $g(x) = x^2$. Solve $f(x) = g(x)$, and explain what your answer means for the two graphs.

Solution

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

So $x = 1$, the only solution. At $x = 1$ both functions equal $1$, so the graphs meet at $(1, 1)$. Since there’s only one solution, the line $y = 2x - 1$ touches the parabola $y = x^2$ at that single point without crossing it (it’s a tangent line).