Week 1 · Day 4 Lesson 1.8–1.9 About 14 minutes Free

Domain and range

By the end: Find the domain of a function from its formula, including composite functions such as $f(2x-1)$ , and find its range with the right method for each type of function.

Understand it

The domain is every input $x$ the function accepts. The range is every output $y$ it produces. Think of the domain as “how far left and right the graph goes” and the range as “how far down and up”.

Finding the domain. Start with the formula and list what is not allowed:

A denominator cannot be $0$ .
An even root such as $\sqrt{\ \ }$ needs a radicand that is $0$ or positive.
A logarithm such as $\ln x$ needs its argument to be positive. (You will meet logarithms properly in Week 7.)
$x^0$ needs $x\ne0$ .

Solve each restriction, then take the intersection: all the conditions must hold together. Write the answer in interval notation. Do not simplify the formula before finding the domain: $\dfrac{x^2-1}{x-1}$ simplifies to $x+1$ , but $x=1$ is still not allowed.

Composite domains. The brackets of $f(\ \cdot\ )$ always hold the input of $f$ , and that input must lie in the domain of $f$ . If the domain of $f$ is $[0,4]$ , then $f(2x-1)$ needs $0\le 2x-1\le4$ , and you solve for $x$ . If instead you are told the domain of $f(x+1)$ is $[1,3]$ , the $x$ values are in $[1,3]$ , so the inside $x+1$ runs over $[2,4]$ , which is the domain of $f$ .

Finding the range. Choose the method by the type of function:

Quadratic. Complete the square to find the vertex. If the vertex’s $x$ is inside the allowed interval, the vertex gives the minimum (opens up) or maximum (opens down), and the endpoints give the other end. If the vertex is outside, only the endpoints matter.
Square root. $\sqrt{u}\ge0$ . Find the range of $u$ first, then take the root.
Fraction with $x$ only in the denominator. $y=\dfrac{k}{g(x)}$ with $k\ne0$ is never $0$ .
Fraction $\dfrac{ax+b}{cx+d}$ . Divide to separate a constant: $y=\dfrac ac+\dfrac{\text{something}}{cx+d}$ . The constant $\dfrac ac$ is the one value that is missed.

Essential formulas

\text{denominator}\ne0,\qquad \text{radicand of an even root}\ge0,\qquad \text{argument of }\ln>0

Solve each restriction on its own, then keep the

x

that satisfy all of them.

f\ \text{has domain}\ [p,q]\ \Rightarrow\ f(g(x))\ \text{needs}\ p\le g(x)\le q

The expression in the brackets is the input of

f

. Solve the double inequality for

x

y=a(x-h)^2+k\quad\Rightarrow\quad\text{vertex }(h,k)

Complete the square to read the vertex. For

a>0

the minimum value is

k

x=h

, if

h

is allowed.

See it

On [0, 3]: minimum 1 at x = 2 (the vertex), maximum 5 at x = 0. The range is [1, 5]. The endpoint (3, 2) is not the maximum.

Worked examples

Worked example 1

Find the domain of $f(x)=\dfrac{\sqrt{x+2}}{x-1}$ .

The square root needs $x+2\ge0$ , so $x\ge-2$ .
The denominator needs $x-1\ne0$ , so $x\ne1$ .
Both must hold. Take $[-2,+\infty)$ and remove $1$ .

Answer: $[-2,1)\cup(1,+\infty)$

Worked example 2

The domain of $f(x)$ is $[0,4]$ . Find the domain of $f(2x-1)$ . Then, if the domain of $f(x+1)$ is $[1,3]$ , find the domain of $f(2x)$ .

The input $2x-1$ must be in $[0,4]$ : $0\le2x-1\le4$ . Add $1$ : $1\le2x\le5$ . Divide by $2$ : $\tfrac12\le x\le\tfrac52$ .
For the second part, $x\in[1,3]$ is the domain of $f(x+1)$ , so the input $x+1$ runs over $[2,4]$ . That is the domain of $f$ .
Now $f(2x)$ needs $2\le2x\le4$ , so $1\le x\le2$ .

Answer: $\left[\tfrac12,\tfrac52\right]$ and $[1,2]$

Worked example 3

Find the range of $f(x)=x^2-4x+5$ on $[0,3]$ .

Complete the square: $x^2-4x+5=(x-2)^2+1$ . The vertex is $(2,1)$ and the parabola opens up.
The vertex’s $x=2$ lies inside $[0,3]$ , so the minimum is $f(2)=1$ .
Compare the endpoints: $f(0)=5$ and $f(3)=9-12+5=2$ . The larger is $5$ , so the maximum is $5$ .

Answer: $[1,5]$

Worked example 4

Find the range of $y=\dfrac{2x+1}{x-3}$ .

Separate a constant: $2x+1=2(x-3)+7$ , so $y=2+\dfrac{7}{x-3}$ .
Since $x\ne3$ and the numerator $7$ is not $0$ , the fraction $\dfrac7{x-3}$ takes every value except $0$ .
So $y$ takes every value except $2+0=2$ .

Answer: $(-\infty,2)\cup(2,+\infty)$

Try it yourself

Work it on paper first. Open a hint only when you are stuck, then compare with the solution.

Find the domain and the range of $f(x)=\sqrt{9-x^2}$ .

Hint 1 of 2

For the domain, the radicand must be $0$ or positive: $9-x^2\ge0$ .

Hint 2 of 2

For the range, first find the range of $9-x^2$ on that domain: its largest value is at $x=0$ .

Show the step-by-step solution

Domain: $9-x^2\ge0$ means $x^2\le9$ , so $-3\le x\le3$ .
On $[-3,3]$ , $9-x^2$ is largest at $x=0$ where it equals $9$ , and it is $0$ at $x=\pm3$ . So $0\le9-x^2\le9$ .
Take the square root: $0\le\sqrt{9-x^2}\le3$ .

Answer: Domain $[-3,3]$ , range $[0,3]$ .

Common mistakes

Simplifying before finding the domain. $\dfrac{x^2-1}{x-1}=x+1$ still excludes $x=1$ .
Finding a quadratic’s range from the endpoints only. On $[0,3]$ the endpoints give $5$ and $2$ , but the minimum is $1$ at the vertex.
Mixing up the input and the output of a composite: the inside of $f(\ \cdot\ )$ must lie in the domain of $f$ .
Writing a range in terms of $x$ . The range is a set of $y$ values.
Forgetting to combine restrictions. Every condition must hold at once, so intersect them.

Exam tips

Write the three restrictions in your head: denominator, even root, logarithm. Check each one.
For a quadratic on an interval, sketch it: is the vertex inside the interval? Then it is the minimum or maximum.
For $f(g(x))$ questions, always write “the inside is in the domain of $f$ ” and solve that.
For $\dfrac{ax+b}{cx+d}$ the missing $y$ value is $\dfrac ac$ . Confirm it with the division.

In short

Domain: no zero denominators, no negative radicands, positive logarithm arguments; intersect the conditions.
Composite: the inside must lie in the domain of the outside function.
Range: complete the square for quadratics, bound the radicand for roots, separate a constant for fractions.

Knowledge check

4 questions on this lesson, marked as you go. Take them in ACE CSCA with a free account: passing them completes the lesson, and your progress is saved on every device.

All of Week 1 is free: the eight lessons, the daily questions and the Week 1 test.

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