JC1 H2 Math Functions: Domain, Inverse & Composite

JC1 · H2 Mathematics · Study guide

Work from the stated domain to inverse branches, inverse graphs and composite ranges. Follow the method through complete question-based applications.

Concepts · worked methods · graphs14 min read · practice linked below

What to look for in a Functions question

Read the domain before working with the formula. It decides which part of the graph to use, which inverse branch to choose, and whether one function can be applied after another.

The domain is the set of permitted inputs. The range is the set of outputs obtained. In the examples below, carry these sets through each step instead of finding them only at the end.

1. Restrict the domain, then find the inverse

Which square-root branch should we use?

Question. Let $f(x)=5-(x-2)^2$ , $x\leq4$ . Explain why $f^{-1}$ does not exist. A new function $g$ has the same rule, with domain $x\leq k$ . Find the greatest value of $k$ for which $g^{-1}$ exists, then define $g^{-1}$ .

Start with the graph. The maximum point is $(2,5)$ , so $R_f=(-\infty,5]$ .

The horizontal line

y=1

cuts the original graph at

(0,1)

and

(4,1)

. Both points belong to the stated domain.

Hence $f$ is not 1-1 and $f^{-1}$ does not exist. To keep the largest domain of the form $x\leq k$ , retain the left branch up to the turning point. Thus $k=2$ .

On $x\leq2$ , every horizontal line cuts the graph at most once. Now rearrange, using this domain to choose the sign:

\begin{aligned}y&=5-(x-2)^2,\\(x-2)^2&=5-y,\\x-2&=-\sqrt{5-y}\quad(x\leq2),\\x&=2-\sqrt{5-y}.\end{aligned}

Replace $y$ by $x$ . The domain of the inverse is the range of $g$ :

\begin{aligned}g^{-1}(x)&=2-\sqrt{5-x},\\D_{g^{-1}}&=R_g=(-\infty,5],\\R_{g^{-1}}&=D_g=(-\infty,2].\end{aligned}

Check. $g(0)=1$ and $g^{-1}(1)=0$ . Taking the positive square-root branch would give outputs greater than 2 when the inverse input is less than 5, outside $D_g$ .

If only an inverse value is required: solve the original function equation. For example, to find $g^{-1}(1)$ , let $t=g^{-1}(1)$ :

\begin{aligned}g(t)&=1,\\5-(t-2)^2&=1,\\t&=0\text{ or }4.\end{aligned}

Since $t\leq2$ , reject 4. Therefore $g^{-1}(1)=0$ . There is no need to find a general inverse formula for this task.

2. Sketch the inverse and the identity composites

Question. For the same function $g(x)=5-(x-2)^2$ , $x\leq2$ , sketch $g$ and $g^{-1}$ . Then sketch $gg^{-1}$ and $g^{-1}g$ , and solve $gg^{-1}(x)=g^{-1}g(x)$ .

Reflect the graph in $y=x$ . Swap the coordinates of each point: the included endpoint $(2,5)$ becomes $(5,2)$ . The point $(0,1)$ becomes $(1,0)$ .

The two curves are reflections in

y=x

. Use equal scales on the axes when drawing the reflection.

For either identity composite the output is $x$ , but the domain is the domain of the function applied first:

\begin{aligned}g^{-1}g(x)&=x,\qquad x\leq2,\\gg^{-1}(x)&=x,\qquad x\leq5.\end{aligned}

Draw only the permitted part of

y=x

. Both lines continue to the lower left; their right endpoints are included.

To solve the equation, both sides must be defined. Their common domain is $( -\infty,2]$ , and both sides equal $x$ there. Hence the solution is $x\leq2$ . The two functions have the same rule but are not equal as functions because their domains differ.

For an inverse graph with an asymptote, swap that too: a vertical asymptote $x=a$ becomes a horizontal asymptote $y=a$ .

3. Check whether the composite exists, then find its range

Use the inner range as the domain of the outer graph

Question. The functions are

\begin{aligned}p(x)&=\left|\frac1{2x-5}\right|,\quad x<\frac52,\\h(x)&=(x-3)^2+1,\quad x\geq-2.\end{aligned}

Explain why $hp$ exists, find its rule and range, and determine whether $ph$ exists.

First use the stated domain to remove the modulus. Since $x<\frac52$ , the expression $2x-5$ is negative. Thus

\begin{aligned}p(x)&=-\frac1{2x-5}\\&=\frac1{5-2x}.\end{aligned}

As $x$ varies over $(-\infty,\frac52)$ , the denominator $5-2x$ takes every positive value. Its reciprocal therefore takes every positive value too, so $R_p=(0,\infty)$ .

In $hp$ , apply $p$ first. Compare its entire range with the domain of $h$ :

\begin{aligned}R_p&=(0,\infty),\\D_h&=[-2,\infty),\\R_p&\subseteq D_h.\end{aligned}

Hence $hp$ exists. Substitute $p(x)$ into $h$ , keeping the domain of $p$ :

\begin{aligned}(hp)(x)&=h\left(\frac1{5-2x}\right)\\&=\left(\frac1{5-2x}-3\right)^2+1\\&=\left(\frac{6x-14}{5-2x}\right)^2+1,\\D_{hp}&=\left(-\infty,\frac52\right).\end{aligned}

To find the range: find $R_p$ , sketch $h$ , and use $R_p$ as the domain of this graph. We therefore look at $h(u)=(u-3)^2+1$ for $u>0$ .

Follow an input through the same composite

Move the value $u=p(x)$ passed into $h$ . Watch the point move down to the minimum and then up again.

Value passed into h:3.0

Retain the green part of

h

. Its turning point

(3,1)

is included because

3\in R_p

. The grey part is not used by the composite.

x = 2.333 → p(x) = 3.0 → h(p(x)) = 1.00. This is the minimum.

The control shows selected positive inputs; the full inner range is all $u>0$ .

The minimum value is 1. It is attained: solve $p(x)=3$ to check the corresponding original input.

\begin{aligned}\frac1{5-2x}&=3,\\1&=15-6x,\\x&=\frac73<\frac52.\end{aligned}

As $u$ increases without bound, $h(u)$ also increases without bound. Hence $R_{hp}=[1,\infty)$ . Looking only at the endpoints of the input interval would miss the minimum at $u=3$ .

Now reverse the order. $R_h=[1,\infty)$ is not contained in $D_p=(-\infty,\frac52)$ . For instance, $h(0)=10$ , but 10 is not a permitted input to $p$ . Hence $ph$ does not exist on the stated domain of $h$ . Being able to calculate some values does not establish that the whole composite exists.

Related task: find the inverse of the modulus branch

Use the same $p$ , with $x<\frac52$ . We have already removed the modulus on this domain:

\begin{aligned}y&=\frac1{5-2x},\\5-2x&=\frac1y,\\x&=\frac52-\frac1{2y}.\end{aligned}

Thus $p^{-1}(x)=\frac52-\frac1{2x}$ , with domain $x>0$ , since $D_{p^{-1}}=R_p$ . The domain makes the inverse output less than $\frac52$ , as required.

Before you finish

For an inverse, did you give the domain and justify the chosen branch?
For a composite, did you compare the inner range with the outer domain in the correct order?
For a range, did you check turning points and whether endpoints are included?
For $ff^{-1}$ or $f^{-1}f$ , did you keep the domain of the function applied first?

Continue with topic questions

Functions practice questions

JC1 H2 Mathematics: Functions, Inverses and Composition