Finding the Range, foundations for calculus lesson

A ball thrown upward never goes higher than some peak, and it never goes below the ground. Those two facts bound its height, but they do not yet say that every height in between is actually reached. Finding a range takes both halves of the argument, a bound that rules values out and a reason that every remaining value really occurs. Calculus leans on the same two halves when it talks about extreme values and the Intermediate Value Theorem.

Key terms

Range

Definition The range of a function $f$ with domain $D$ is the set of all outputs it produces,

\{\, f(x) \mid x \in D \,\}.

The range depends on both the formula and the domain.

In plain words Feed the function every allowed input and collect every answer it gives. That collection is the range.

Example Let $f(x) = x^2$ on the domain $\{-1, 0, 1, 2\}$ . The outputs are $1$ , $0$ , $1$ and $4$ , so the range is $\{0, 1, 4\}$ . The repeated output $1$ is listed once.

Attained output

Definition A number $y$ is an attained output of $f$ if there is at least one $x$ in the domain with $f(x) = y$ . To prove that $y$ belongs to the range, produce such an $x$ and check that it lies in the domain. A bound such as $f(x) \ge 0$ for every $x$ only shows that the range is contained in $[0, \infty)$ . It does not show that every number in that interval is attained.

In plain words To show a number is an output, name an input that gives it. Knowing where the outputs cannot be is only half the job.

Example Take $f(x) = x^2 + 1$ with domain all real numbers. Since $x^2 \ge 0$ , every output satisfies $f(x) \ge 1$ . Now pick any $y \ge 1$ and choose $x = \sqrt{y - 1}$ , which is a real number because $y - 1 \ge 0$ . Then

f(\sqrt{y - 1}) = (y - 1) + 1 = y.

So every $y \ge 1$ is attained, and the range is $[1, \infty)$ . For contrast, the outputs of $x^2$ on the integers are also all at least $0$ , yet $2$ is never attained, since no integer squares to $2$ .

Domain restriction

Definition Restricting a function to a smaller domain means keeping the same formula but allowing only some of the original inputs. The range of the restricted function is the set of outputs of those inputs alone, so it can be smaller than the original range, even though the formula is unchanged.

In plain words Fewer inputs can mean fewer outputs. Always find the range for the domain you were actually given.

Example The formula $x^2$ has range $[0, \infty)$ on all real numbers. On the domain $[1, 3]$ the function increases from $1$ to $9$ , so its range is $[1, 9]$ . On $[-1, 2]$ the inputs pass through $0$ , so the range starts at $0$ . Every output satisfies $x^2 \le 4$ there, because $\lvert x \rvert \le 2$ . Every $y$ in $[0, 4]$ is attained by $x = \sqrt{y}$ , which lies in $[0, 2]$ . The range on $[-1, 2]$ is $[0, 4]$ .

Minimum and maximum

Definition A function $f$ with domain $D$ has a minimum value $f(c)$ at $c \in D$ if $f(c) \le f(x)$ for every $x \in D$ , and a maximum value $f(d)$ at $d \in D$ if $f(d) \ge f(x)$ for every $x \in D$ . Both are outputs that are actually attained. A lower bound that is never attained is not a minimum, and an upper bound that is never attained is not a maximum.

In plain words The minimum is the smallest output the function really reaches, and the maximum is the largest. If the function only creeps toward a value without reaching it, that value is not a minimum or maximum.

Example On $[-1, 2]$ the function $x^2$ has minimum $0$ at the interior point $x = 0$ and maximum $4$ at the endpoint $x = 2$ . The other endpoint, $x = -1$ , gives $1$ , which is neither. So an extreme value can sit at an endpoint or inside the interval, and you have to check both.

Endpoint inclusion

Definition When a range is an interval, an endpoint of that interval belongs to the range exactly when some input in the domain produces it. So the endpoints of the domain need a check of their own. An input at an included endpoint gives an output that is attained, while a value that the outputs only approach as $x$ nears an excluded endpoint may be missing from the range.

In plain words A bracket in the range means some input really hits that value. A parenthesis at a finite end means the outputs get arbitrarily close to that value but never arrive.

Example Let $f(x) = x^2$ on $(0, 2]$ . Every output satisfies $0 < x^2 \le 4$ . The value $4$ is attained at the included endpoint $x = 2$ , but $0$ would need $x = 0$ , which is not in the domain. Every $y$ with $0 < y \le 4$ is attained by $x = \sqrt{y}$ , which lies in $(0, 2]$ . The range is $(0, 4]$ , and this function has a maximum but no minimum.

Output interval

Definition An output interval is a range written as an interval of $y$ values, for example $[k, \infty)$ or $(-\infty, M]$ . To establish it, show two things. First, every output lies in the interval. Second, every number in the interval is attained by some input in the domain. When $f$ is continuous on an interval, its range on that interval is always an interval, possibly a single point if $f$ is constant. This fact comes from the Intermediate Value Theorem in calculus. Without continuity, or on a domain with gaps, the range can have gaps too. Even when the theorem applies, the examples here name the inputs directly.

In plain words On a graph, the range is the shadow the curve casts on the $y$ axis. You find its edges with bounds, then show the shadow has no gaps by naming inputs.

Example Find the range of the decreasing function $f(x) = 3 - \sqrt{x}$ on $[0, \infty)$ . Since $\sqrt{x} \ge 0$ , every output satisfies $f(x) \le 3$ . Now take any $y \le 3$ and choose $x = (3 - y)^2$ , which is at least $0$ and so lies in the domain. Then

\begin{aligned} f\big((3 - y)^2\big) &= 3 - \sqrt{(3 - y)^2} \\[0.4em] &= 3 - \lvert 3 - y \rvert \\[0.4em] &= 3 - (3 - y) = y. \end{aligned}

The step $\lvert 3 - y \rvert = 3 - y$ is valid because $y \le 3$ . The range is $(-\infty, 3]$ , with maximum $3$ at $x = 0$ and no minimum. For instance $y = -5$ comes from $x = 64$ , and indeed $3 - \sqrt{64} = -5$ .

Check yourself

Find the range of $f(x) = x^2$ on $[-3, 1]$ , and give its minimum and maximum.

Show the answer

On this domain $\lvert x \rvert \le 3$ , so $0 \le x^2 \le 9$ . Every $y$ in $[0, 9]$ is attained by $x = -\sqrt{y}$ , which lies in $[-3, 0]$ . The range is $[0, 9]$ , with minimum $0$ at $x = 0$ and maximum $9$ at the endpoint $x = -3$ .

Find the range of $g(x) = 2 - \sqrt{x - 1}$ on $[1, \infty)$ .

Show the answer

Since $\sqrt{x - 1} \ge 0$ , every output is at most $2$ . For any $y \le 2$ choose $x = 1 + (2 - y)^2$ , which is at least $1$ . Then

\begin{aligned} g(x) &= 2 - \sqrt{(2 - y)^2} \\[0.4em] &= 2 - \lvert 2 - y \rvert \\[0.4em] &= 2 - (2 - y) = y, \end{aligned}

using $y \le 2$ . The range is $(-\infty, 2]$ .

Find the range of $h(x) = \frac{1}{x}$ on $[1, \infty)$ .

Show the answer

For $x \ge 1$ the output satisfies $0 < \frac{1}{x} \le 1$ . Each $y$ with $0 < y \le 1$ is attained by $x = \frac{1}{y}$ , which is at least $1$ . The value $0$ is never attained. The range is $(0, 1]$ .

Finding the Range

Key terms

Range

Attained output

Domain restriction

Minimum and maximum

Endpoint inclusion

Output interval

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus