courses › Algebra

Functions

level 27 course

Composition, inverses, domains, and piecewise rules.

Pen and paper is fine · no calculator needed why?

Opens at level 27.

Builds on: Lines & slope (not open yet)

the lesson

The idea, the techniques and a tip for each skill, right here. The Learn page adds worked examples for every skill and untimed practice.

Read the lesson · about 3 minutes

The idea

A function is a rule that turns each input into one output: f(x) = 2x + 3 turns 4 into 11.

A piecewise function uses different rules on different stretches of x. A composition, like f(g(3)), feeds one function's output into another. An inverse, written f⁻¹, runs the rule backward: f⁻¹(11) asks which input gave 11. The domain is every input the rule can accept.

Techniques

Pick the piece first

The function has different rules for different values of x.

For each input, check which condition it meets.
At the boundary, look for ≤ or ≥. The piece with the bar under its sign includes the boundary.
Use only that piece's rule for that input.

worked example

Example: f(x) = x² when x < 1, and 2x + 5 when x ≥ 1. What is f(1) + f(−2)?

1 meets x ≥ 1, so f(1) = 2 × 1 + 5 = 7.
−2 meets x < 1, so f(−2) = (−2)² = 4.
Total: 7 + 4 = 11.

Answer: 11

Work inside out

One function sits inside another, like f(g(2)).

Work out the inside function first.
Feed that number into the outside function.
f(g(2)) and g(f(2)) are usually different, so read the order.

worked example

Example: f(x) = 3x − 1 and g(x) = x² + 2. What is f(g(2))?

Inside first: g(2) = 2² + 2 = 6.
Then f(6) = 3 × 6 − 1 = 17.

Answer: 17

What the rule allows

You need the domain: the inputs a function accepts.

Square root: the inside must be 0 or more.
Log: the inside must be more than 0.
Fraction: the bottom cannot be 0.
Solve that condition for x.

worked example

Example: What is the smallest integer x in the domain of f(x) = log(x − 2)?

A log needs a positive inside: x − 2 > 0.
So x > 2, which leaves out 2 itself.
The smallest integer above 2 is 3.

Answer: 3

Tips by skill

TipPiecewise functions: For each input, find which condition it meets before you use a rule. At the boundary, the piece with ≤ or ≥ owns it.
TipComposition: Work inside out: find the inner function's value first, then feed it to the outer one.
TipInverse value: f⁻¹(y) is the input that gives y. Set the rule equal to y and solve for x.
TipDomain: Square root: inside at least 0. Log: inside more than 0. Fraction: bottom not 0. Solve that for x.

Watch out for

Using the wrong piece at the boundary. Check which condition has ≤ or ≥.
Working f(g(3)) from the outside in. Start with the inside function.
Working out f(8) when asked for f⁻¹(8). The inverse asks which input gives 8.
Leaving the boundary out of a square root's domain. The square root of 0 is allowed; it is a log that needs more than 0.

skills · practice stats

From rounds of this course only: box, review and test-out answers are left out. Once a skill has 40 tries, it compares your first 20 tries with your last 20.

Piecewise functions par 18 s not tried yet
worked example
f(x) = 2x + 8 when x ≤ 0, and x² when x > 0. What is f(0) + f(5)?

Answer: 33
1. f(0) uses the first rule (x ≤ 0): 2 × 0 + 8 = 0 + 8 = 8.
2. f(5) uses the second rule (x > 0): 5² = 25.
3. Total: 8 + 25 = 33.
Composition par 20 s not tried yet
worked example
f(x) = 2x + 6 and g(x) = x − 4. What is g(f(2))?

Answer: 6
1. Work inside out: f(2) = 2 × 2 + 6 = 4 + 6 = 10.
2. Then g(10) = 10 − 4 = 6.
Inverse value par 20 s not tried yet
worked example
f(x) = 3x − 10. What is f⁻¹(−31)?

Answer: -7
1. f⁻¹(−31) is the input that gives −31: solve 3x − 10 = −31.
2. 3x = −21, so x = −7.
Domain par 16 s not tried yet
worked example
What is the domain of f(x) = √(−2 − x)?
1. x ≤ 2
2. all real x
3. x < −2
4. x ≥ −2
5. x ≤ −2
Answer: x ≤ −2
1. A square root needs a non-negative inside: −2 − x ≥ 0, so x ≤ −2.

rest ladder

Win 3 of your last 4 rounds and the course rests. A win is 90% right, within 2× the round's par. Pass the review when it comes back and the next rest is longer.

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