The Range of a Function | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Find the lowest and highest outputs

Sketch or inspect the graph only over its stated domain.
Check both domain boundaries and every turning point inside it.
Decide whether each boundary output is included, then write the answer using $\mathrm{f}(x)$ or $y$ .

Here is $\mathrm{f}(x)=(x-3)^2+8$ on the domain $-1\le x\le 9$ . Its vertex (turning point) $(3,8)$ is inside the domain, so it gives the minimum. The endpoint $x=9$ gives the maximum $44$ .

The range is the gold vertical line: lowest output 8 (the vertex) up to highest 44 (the end point x = 9).

Answer

8\le \mathrm{f}(x)\le 44

Open endpoints need one extra check

For a straight line on a closed interval, testing both endpoints is enough. For a curve, also test any turning point inside the interval. With an open endpoint, an excluded output may still be produced by a different allowed input.

Worked example

State the range of $\mathrm{g}(x)=(x+1)^2-4$ for $-3<x\le2$

The vertex is at $x=-1$ , inside the domain, so the minimum $-4$ is included.
Test the two domain boundaries. At $x=-3$ the output would be $0$ , but $x=-3$ is excluded. At $x=2$ , the included output is $5$ .
The maximum is $5$ . Although $x=-3$ is excluded, its output $0$ remains in the range because the allowed input $x=1$ also gives $0$ .

Answer

-4\le \mathrm{g}(x)\le5

Two range checks

Common mistake

Write a range using

\mathrm{f}(x)

(or

y

), not an interval of input values. The mark scheme rejects an

x

-interval as a range. (9709/12 Oct/Nov 2021 Q10)

Common mistake

A turning point affects the range only when its

x

-value lies inside the domain. Otherwise use the allowed boundary values.

Now you try

Your turntextbook Explore 2.2-style

The function $\mathrm{f}(x)=(x-2)^2+1$ is defined for $x\in\mathbb{R}$ . Find the range of $\mathrm{f}$ , and say whether it is one-one or many-one. (textbook Explore 2.2-style)

Show worked answer

The squared bracket is never negative, so its smallest value is $0$ , at the vertex $x=2$ , which makes the lowest output $0+1=1$ , and from there the curve goes up with no top.

Answer

\mathrm{f}(x)\ge 1

Over all of $\mathbb{R}$ , the two sides of the parabola give the same outputs. So it is many-one. That is why you would need a smaller domain before you could find an inverse.

Find the lowest and highest outputs

Sketch or inspect the graph only over its stated domain.
Check both domain boundaries and every turning point inside it.
Decide whether each boundary output is included, then write the answer using $\mathrm{f}(x)$ or $y$ .

Here is $\mathrm{f}(x)=(x-3)^2+8$ on the domain $-1\le x\le 9$ . Its vertex (turning point) $(3,8)$ is inside the domain, so it gives the minimum. The endpoint $x=9$ gives the maximum $44$ .

The range is the gold vertical line: lowest output 8 (the vertex) up to highest 44 (the end point x = 9).

Answer

8\le \mathrm{f}(x)\le 44

Open endpoints need one extra check

Worked example

State the range of $\mathrm{g}(x)=(x+1)^2-4$ for $-3<x\le2$

The vertex is at $x=-1$ , inside the domain, so the minimum $-4$ is included.
Test the two domain boundaries. At $x=-3$ the output would be $0$ , but $x=-3$ is excluded. At $x=2$ , the included output is $5$ .
The maximum is $5$ . Although $x=-3$ is excluded, its output $0$ remains in the range because the allowed input $x=1$ also gives $0$ .

Answer

-4\le \mathrm{g}(x)\le5

Two range checks

Common mistake

Write a range using

\mathrm{f}(x)

(or

y

), not an interval of input values. The mark scheme rejects an

x

-interval as a range. (9709/12 Oct/Nov 2021 Q10)

Common mistake

A turning point affects the range only when its

x

-value lies inside the domain. Otherwise use the allowed boundary values.

Now you try

Your turntextbook Explore 2.2-style

The function $\mathrm{f}(x)=(x-2)^2+1$ is defined for $x\in\mathbb{R}$ . Find the range of $\mathrm{f}$ , and say whether it is one-one or many-one. (textbook Explore 2.2-style)

Show worked answer

The squared bracket is never negative, so its smallest value is $0$ , at the vertex $x=2$ , which makes the lowest output $0+1=1$ , and from there the curve goes up with no top.

Answer

\mathrm{f}(x)\ge 1

Over all of $\mathbb{R}$ , the two sides of the parabola give the same outputs. So it is many-one. That is why you would need a smaller domain before you could find an inverse.