Composite and Inverse Functions

Worked example 1 — evaluate in both orders

Let \(f(x)=2x+1\) and \(g(x)=x^2\).

1. Find \((f\circ g)(3)\).

2. Find \((g\circ f)(3)\).

Solution

(a) \(g(3)=3^2=9\), so \(f(g(3))=2(9)+1=\boxed{19}\).

(b) \(f(3)=2(3)+1=7\), so \(g(f(3))=7^2=\boxed{49}\).

The same two functions give different answers when their order is reversed.

Worked example 2 — replace every \(x\) by \(g(x)\)

Let \(f(x)=2x^2-5x+7\) and \(g(x)=3x-1\).

1. Find an expression for \((f\circ g)(x)\).

2. Find an expression for \((g\circ f)(x)\).

3. Find \((f\circ g)(2)\).

Solution

(a) First write the rule using \(g(x)\). Replace both occurrences of \(x\) in \(f(x)=2x^2-5x+7\) by \(g(x)\): \[f(g(x))=2[g(x)]^2-5g(x)+7.\] Then substitute the expression for \(g(x)\). Since \(g(x)=3x-1\), \[\begin{aligned} f(g(x))&=2(3x-1)^2-5(3x-1)+7\\ &=18x^2-12x+2-15x+5+7\\ &=\boxed{18x^2-27x+14}. \end{aligned}\] (b) In the other order, replace \(x\) in \(g(x)=3x-1\) by \(f(x)\): \[\begin{aligned} g(f(x))&=3f(x)-1\\ &=3(2x^2-5x+7)-1\\ &=\boxed{6x^2-15x+20}. \end{aligned}\] (c) Using the expression from part (a), \[\begin{aligned}(f\circ g)(2)&=18(2)^2-27(2)+14\\&=\boxed{32}\end{aligned}\].

Check by stages: \(g(2)=5\), then \[\begin{aligned}f(5)&=2(5)^2-5(5)+7\\&=32\end{aligned}\].

Worked example 3 — a formula is not always needed

Use the table to answer the following questions.

Solution

\[\begin{aligned} \textbf{(a)}\quad f(g(0))&=f(1)=\boxed{3},\\ \textbf{(b)}\quad g(f(0))&=g(2)=\boxed{3},\\ \textbf{(c)}\quad f(f(1))&=f(3)=\boxed{0}. \end{aligned}\] The first two answers happen to agree at \(x=0\). This does not prove that the two composite functions are equal for every input.

Follow one input — then find all possible final outputs

Let \(g(x)=x+1\), \(0\leq x\leq2\), and \(f(x)=x^2\), \(x\in\mathbb{R}\).

1. Follow the input \(a=1\) through \(g\), then through \(f\).

2. Find the domain and range of \(f\circ g\).

Solution

\[\textbf{(a)}\quad\underbrace{1}_{a}\xrightarrow{\ g\ }\underbrace{2}_{g(a)} \xrightarrow{\ f\ }\underbrace{4}_{f(g(a))}.\] (b) For all starting inputs, \(g\) sends \([0,2]\) to \([1,3]\), and \(f\) squares these positive intermediate values to give \([1,9]\). Therefore \[\boxed{\begin{gathered}(f\circ g)(x)=(x+1)^2\\\operatorname{Dom}(f\circ g)=[0,2]\\\operatorname{Ran}(f\circ g)=[1,9].\end{gathered}}\] Here \(\operatorname{Ran} f=[0,\infty)\), but the composite only reaches \([1,9]\).

See the two stages. The middle value \(u=g(x)\) is the horizontal input in the sketch of \(f\). Filled endpoints show that the interval boundaries are included.

Stage 1: \(g(x)=x+1\)

\(\operatorname{Dom} g=[0,2]\), \(\operatorname{Ran} g=[1,3]\).
The marked interior point shows \(g(1)=2\).

Stage 2: \(f(u)=u^2\)

Full function: \(\operatorname{Dom} f=\mathbb{R}\), \(\operatorname{Ran} f=[0,\infty)\).
Solid part: inputs \([1,3]\), outputs \([1,9]\).

The composite: \((f\circ g)(x)=(x+1)^2\)

Read the domain horizontally.

The composite starts at \(x=0\) and ends at \(x=2\): its domain is \([0,2]\).

Read the range vertically.

Its lowest output is \(1\), and its highest is \(9\). Every value between them is reached: its range is \([1,9]\).

Why only part of the parabola?

Although \(f\) accepts every real input, \(g\) supplies only values from \(1\) to \(3\). Only the solid part of the sketch of \(f\) is used.

\[\begin{gathered}\text{Starting inputs }[0,2]\\\downarrow\ g\\\text{Middle values }[1,3]\\\downarrow\ f\\\text{Final outputs }[1,9]\end{gathered}\]

Worked example 4 — reversing the order changes the domain

Let \(f(x)=\sqrt{x}\), \(x\geq0\), and \(g(x)=x-3\), \(x\in\mathbb{R}\).

1. Find \((f\circ g)(x)\) and state its domain.

2. Find \((g\circ f)(x)\) and state its domain.

Solution

(a) The first stage \(g\) accepts every real number, but \(f\) requires its input \(g(x)=x-3\) to be non-negative. Thus \(x-3\geq0\), giving \[\boxed{\begin{gathered}(f\circ g)(x)=\sqrt{x-3}\\x\geq3.\end{gathered}}\]

(b) In the reverse order, \(g(f(x))=\sqrt{x}-3\). The square root requires \(x\geq0\), and subtracting 3 adds no restriction: \[\boxed{\begin{gathered}(g\circ f)(x)=\sqrt{x}-3\\x\geq0.\end{gathered}}\] For example, \(x=1\) is allowed in \(g\circ f\), but not in \(f\circ g\).

Worked example 5 — simplification can hide an excluded input

Let \(f(x)=1/x\), \(x\ne0\), and \(g(x)=1/(x-2)\), \(x\ne2\).

1. Find the domain of \(f\circ g\).

2. Find and simplify an expression for \((f\circ g)(x)\).

Solution

(a) The inner function requires \(x\ne2\). Its output \(1/(x-2)\) is never zero, so it is always accepted by \(f\). Thus \(\boxed{\operatorname{Dom}(f\circ g)=\mathbb{R}\setminus\{2\}}\).

(b) Substitute \(g(x)\) into \(f\), then simplify: \[f(g(x))=\frac{1}{\frac{1}{x-2}}=\boxed{x-2,\quad x\ne2}.\] Although the simplified expression \(x-2\) can be evaluated at 2, the original composition cannot: \(g(2)\) is undefined.

Worked example 6 — two possible inner functions

The functions \(f\) and \(g\) are defined by \[f(x)=ax+b,\qquad g(x)=x^2-5x+6,\qquad a,b\in\mathbb{R}.\] Given that \((g\circ f)(x)=4x^2-6x+2\) for all \(x\in\mathbb{R}\), find both possible expressions for \(f(x)\).

Solution

First substitute \(f(x)\) into \(g\): \[\begin{aligned} g(f(x))&=(ax+b)^2-5(ax+b)+6\\ &=a^2x^2+a(2b-5)x+(b^2-5b+6). \end{aligned}\] Since the expressions agree for every real \(x\), their corresponding coefficients are equal: \[\begin{array}{c|c|c} \text{coefficient of }x^2&\text{coefficient of }x&\text{constant term}\\[2pt] \boxed{a^2=4}&\boxed{a(2b-5)=-6}&\boxed{b^2-5b+6=2} \end{array}\] The first equation gives \(a=2\) or \(a=-2\). Keep both possibilities.

Hence the two possible functions are \[\boxed{f(x)=2x+1\quad\text{or}\quad f(x)=-2x+4.}\]

Worked example 7 — reverse the order and the operations

Let \(f(x)=3x-5\), with domain \([1,4]\).

1. Find an expression for \(f^{-1}(x)\).

2. State the domain and range of \(f^{-1}\).

3. Find \(f^{-1}(7)\) and explain its meaning.

Solution

(a) Reverse the order and undo each operation:

Therefore \(\boxed{f^{-1}(x)=\dfrac{x+5}{3}}\).

(b) Since \(f\) is increasing, its endpoint values give its range: \[f(1)=-2,\qquad f(4)=7\quad\Longrightarrow\quad \operatorname{Ran} f=[-2,7].\] Thus \(\operatorname{Dom} f^{-1}=[-2,7]\) and \(\operatorname{Ran} f^{-1}=[1,4]\).

(c) \(f^{-1}(7)=4\) means that the input which produces an output of 7 is 4.

Worked example 8 — restrict a domain to obtain an inverse

Consider \(f(x)=x^2\), initially with domain \(\mathbb{R}\).

1. Explain why \(f\) has no inverse function on this domain.

2. Give a domain restriction for which an inverse exists, and find that inverse.

3. Give a different domain restriction and find the corresponding inverse.

Solution

(a) The output 4 comes from both \(-2\) and 2. Reversing this would require the inverse to send 4 to two different outputs.

(b) Restrict to \(x\geq0\): each non-negative output comes from one non-negative input. Thus \[\boxed{f^{-1}(x)=\sqrt{x},\ x\geq0.}\]

(c) Restrict instead to \(x\leq0\): each non-negative output comes from one non-positive input. Thus \[\boxed{f^{-1}(x)=-\sqrt{x},\ x\geq0.}\]

Worked example 9 — a cubic function

Let \(f(x)=(x-2)^3+1\), \(x\in\mathbb{R}\).

1. Find an expression for \(f^{-1}(x)\).

2. State the domain and range of \(f^{-1}\).

Solution

(a) This function is one-to-one. Start with \(y=(x-2)^3+1\): \[\begin{aligned} x&=(y-2)^3+1 &&\text{swap }x\text{ and }y,\\ x-1&=(y-2)^3 &&\text{subtract 1},\\ \sqrt[3]{x-1}&=y-2 &&\text{take the cube root},\\ y&=2+\sqrt[3]{x-1} &&\text{add 2}. \end{aligned}\] Thus \(\boxed{f^{-1}(x)=2+\sqrt[3]{x-1}}\). A real cube root has a unique real value; there is no \(\pm\) choice.

(b) The domain and range of \(f\) are both \(\mathbb{R}\), so \[\boxed{\begin{gathered}\operatorname{Dom} f^{-1}=\mathbb{R}\\\operatorname{Ran} f^{-1}=\mathbb{R}\end{gathered}}\].

Worked example 10 — a rational function

Let \(g(x)=\dfrac{2x+3}{x-1}\), \(x\ne1\).

1. Find an expression for \(g^{-1}(x)\).

2. State the domain and range of \(g^{-1}\).

Solution

(a) Swap the variables and rearrange: \[\begin{aligned} x&=\frac{2y+3}{y-1} &&\Longrightarrow& x(y-1)&=2y+3\\ xy-x&=2y+3 &&\Longrightarrow& y(x-2)&=x+3. \end{aligned}\] Therefore \[\boxed{g^{-1}(x)=\frac{x+3}{x-2}.}\] Check a pair: \(g(2)=7\), and \[\begin{aligned}g^{-1}(7)&=\frac{7+3}{7-2}\\&=2\end{aligned}\].

(b) Writing \(g(x)=2+5/(x-1)\) shows that its range is \(\mathbb{R}\setminus\{2\}\); every value other than 2 occurs exactly once. Swapping domain and range gives \[\boxed{\begin{gathered}\operatorname{Dom} g^{-1}=\mathbb{R}\setminus\{2\}\\\operatorname{Ran} g^{-1}=\mathbb{R}\setminus\{1\}.\end{gathered}}\]

Worked example 11 — the same quadratic, two domain choices

Find each inverse, stating its domain and range.

1. \(f(x)=x^2-4x+7\), \(x\geq2\).

2. \(g(x)=x^2-4x+7\), \(x\leq2\).

Solution

For both parts, complete the square: \[\begin{aligned}x^2-4x+7&=(x-2)^2+3\end{aligned}\]. Both restricted functions have range \([3,\infty)\). After swapping \(x\) and \(y\), \[x=(y-2)^2+3\quad\Longrightarrow\quad(y-2)^2=x-3 \quad\Longrightarrow\quad y=2\pm\sqrt{x-3}.\] This is a choice of two branches, not yet a single inverse function.

Solid: \(f\). Dashed curved branch: \(f^{-1}\).

Solid: \(g\). Dashed curved branch: \(g^{-1}\).

In each sketch, the pale diagonal is \(y=x\). The vertex \((2,3)\) becomes \((3,2)\).

Worked example 12 — a square root and its inverse

Let \(f(x)=\sqrt{x-1}+2\), \(x\geq1\).

1. Find an expression for \(f^{-1}(x)\).

2. State the domain and range of \(f^{-1}\).

3. Verify that \(f^{-1}(f(x))=x\), stating the domain on which this holds.

4. Verify that \(f(f^{-1}(x))=x\), stating the domain on which this holds.

Solution

(a) The square root is non-negative, so \(\operatorname{Ran} f=[2,\infty)\). Swapping \(x\) and \(y\), \[\begin{aligned} x&=\sqrt{y-1}+2,\\ x-2&=\sqrt{y-1} &&\text{so }x\geq2,\\ (x-2)^2&=y-1,\\ y&=(x-2)^2+1. \end{aligned}\] Thus \(\boxed{f^{-1}(x)=(x-2)^2+1}\).

(b) Swap the domain and range of \(f\): \[\boxed{\begin{gathered}\operatorname{Dom} f^{-1}=[2,\infty)\\\operatorname{Ran} f^{-1}=[1,\infty)\end{gathered}}\].

(c) For \(x\geq1\), \[\begin{aligned} f^{-1}(f(x)) &=\left(\sqrt{x-1}+2-2\right)^2+1\\ &=(x-1)+1=\boxed{x}. \end{aligned}\] (d) In the reverse order, for \(x\geq2\), \[\begin{aligned} f(f^{-1}(x)) &=\sqrt{(x-2)^2+1-1}+2\\ &=\sqrt{(x-2)^2}+2\\ &=|x-2|+2\\ &=(x-2)+2=\boxed{x},\qquad\text{because }x\geq2. \end{aligned}\] Both compositions give the identity, each on its appropriate domain.

Worked example 13 — a sequence of sketches

The graph of \(f(x)=x^3+1\), \(-1\leq x\leq1\), passes through \((-1,0)\), \((0,1)\) and \((1,2)\).

1. Sketch \(f^{-1}\), showing the corresponding points.

2. State the domain and range of \(f^{-1}\).

Solution

(a) Swap the coordinates of each point, then reflect the curve as shown below.

1. Mark the original points

2. Swap the coordinates

3. Reflect the curve

Worked example 14 — sketch first, then confirm algebraically

Let \(f(x)=1+\dfrac2{x-3}\), \(x\ne3\).

1. Sketch \(f^{-1}\), using asymptotes and points.

2. Find an expression for \(f^{-1}(x)\).

3. State the domain and range of \(f^{-1}\).

Solution

(a) Swap horizontal and vertical asymptotes, and reverse each point’s coordinates.

Original graph

Reflected graph

(b) Swap \(x,y\): \(x=1+\dfrac2{y-3}\). Then \(y-3=\dfrac2{x-1}\), so \[\boxed{f^{-1}(x)=3+\frac2{x-1}.}\] (c) The original domain excludes 3 and the range excludes 1. Hence \[\boxed{\begin{gathered}\operatorname{Dom} f^{-1}=\mathbb{R}\setminus\{1\}\\\operatorname{Ran} f^{-1}=\mathbb{R}\setminus\{3\}.\end{gathered}}\]

Worked example 15 — a decreasing square-root function

Let \(f(x)=\sqrt{6-2x}-1\) on its largest real domain.

1. State the domain and range of \(f\).

2. Find \(f^{-1}\), giving its domain and range.

3. Sketch \(f\) and \(f^{-1}\) on the same axes.

Solution

(a) Domain and range of \(f\). The radicand must be non-negative: \[6-2x\geq0\quad\Longrightarrow\quad x\leq3.\] The square root is at least 0 and grows without bound as \(x\) decreases. Therefore \[\operatorname{Dom} f=(-\infty,3],\qquad\operatorname{Ran} f=[-1,\infty).\] The function is strictly decreasing, hence one-to-one.

(b) Find the inverse. Swap \(x\) and \(y\), then rearrange: \[\begin{aligned} x&=\sqrt{6-2y}-1,\\ x+1&=\sqrt{6-2y} &&\text{requires }x\geq-1,\\ (x+1)^2&=6-2y,\\ y&=3-\frac{(x+1)^2}{2}. \end{aligned}\] Thus \[\boxed{\begin{gathered}f^{-1}(x)=3-\frac{(x+1)^2}{2}\\x\geq-1\\\operatorname{Ran} f^{-1}=(-\infty,3].\end{gathered}}\]

(c) Sketches of \(f\) and \(f^{-1}\)

The endpoint \((3,-1)\) becomes \((-1,3)\).

The point \((1,1)\) lies on the mirror line, so it stays fixed.

The inverse is only the branch with \(x\geq-1\), not a complete parabola.

Solid: \(f\). Dashed curve: \(f^{-1}\). Pale diagonal: \(y=x\).

Worked example 16 — show that a function is self-inverse

Show that \(f(x)=\dfrac{x+3}{x-1}\), \(x\ne1\), is self-inverse.

Solution

Method 1: find the inverse. Start with \(y=(x+3)/(x-1)\) and swap the variables: \[\begin{aligned} x&=\frac{y+3}{y-1}\\ xy-x&=y+3\\ y(x-1)&=x+3\\ y&=\frac{x+3}{x-1}. \end{aligned}\] Since \(f(x)=1+4/(x-1)\), its range is \(\mathbb{R}\setminus\{1\}\). Thus the inverse has the same formula and the same domain as \(f\): \(\boxed{f^{-1}=f}\).

Method 2: compose the function with itself. For \(x\ne1\), \[f(f(x))= \frac{\frac{x+3}{x-1}+3}{\frac{x+3}{x-1}-1} =\frac{\frac{4x}{x-1}}{\frac4{x-1}} =\boxed{x}.\] The inner output is never 1, so the second application is defined.

The graph reflects onto itself. Its horizontal and vertical asymptotes are \(y=1\) and \(x=1\), which swap.

For example, \(f(2)=5\) and \(f(5)=2\).

A self-inverse function need not be the identity: it may interchange two different numbers.