Composition, Inverses and Mixed Practice — Worked Solutions

Ques­tion 1: Com­pos­ing func­tions given as sets of pairs#

The prob­lem#

$f = \{(1, 3), (2, 5), (3, 1)\}$ and $g = \{(1, 2), (3, 4), (5, 6)\}$. Find $g \circ f$.

Under­stand­ing the prob­lem#

Each pair $(a, b)$ in $f$ means $f(a) = b$. So $f(1) = 3$, $f(2) = 5$, $f(3) = 1$, and $g(1) = 2$, $g(3) = 4$, $g(5) = 6$. The domain of $g \circ f$ is the domain of $f$: $\{1, 2, 3\}$.

The idea#

For each $x$ in the domain of $f$, find $f(x)$, then feed that into $g$.

Step-by-step solu­tion#

Step 1. $x = 1$: $f(1) = 3$, then $g(3) = 4$. So $(g \circ f)(1) = 4$.

Step 2. $x = 2$: $f(2) = 5$, then $g(5) = 6$. So $(g \circ f)(2) = 6$.

Step 3. $x = 3$: $f(3) = 1$, then $g(1) = 2$. So $(g \circ f)(3) = 2$.

Step 4. Write the result as a set of pairs.

$$g \circ f = \{(1, 4), (2, 6), (3, 2)\}$$

Check­ing the answer#

Every out­put of $f$ ($3$, $5$, $1$) is in the domain of $g$, so $g \circ f$ is defined at every point.

Answer#

$g \circ f = \{(1, 4), (2, 6), (3, 2)\}$

Ques­tion 2: Com­pos­ites of for­mu­las#

The prob­lem#

$f(x) = x + 3$, $g(x) = x^2 - 2$. Find $f \circ g$, $g \circ f$ and $f \circ f$.

Under­stand­ing the prob­lem#

$(f \circ g)(x) = f(g(x))$: put $g(x)$ into $f$. $(g \circ f)(x) = g(f(x))$: put $f(x)$ into $g$.

The idea#

Replace the input of the outer func­tion by the whole inner expres­sion, then sim­plify.

Step-by-step solu­tion#

Step 1. $f \circ g$: $f$ adds $3$ to its input.

$$(f \circ g)(x) = f(x^2 - 2) = (x^2 - 2) + 3 = x^2 + 1$$

Step 2. $g \circ f$: $g$ squares its input and sub­tracts $2$.

$$\begin{aligned} (g \circ f)(x) = g(x + 3) &= (x + 3)^2 - 2\\ &= x^2 + 6x + 9 - 2 = x^2 + 6x + 7 \end{aligned}$$

Step 3. $f \circ f$: add $3$, then add $3$ again.

$$(f \circ f)(x) = f(x + 3) = x + 6$$

Check­ing the answer#

At $x = 1$: $g(1) = -1$, $f(-1) = 2$, and $1^2 + 1 = 2$. Also $f(1) = 4$, $g(4) = 14$, and $1 + 6 + 7 = 14$.

Answer#

$f \circ g = x^2 + 1$; $g \circ f = x^2 + 6x + 7$; $f \circ f = x + 6$.

Com­mon mis­take to avoid#

Mix­ing up the order. $f \circ g$ and $g \circ f$ are dif­fer­ent here.

Ques­tion 3: A func­tion that is its own inverse#

The prob­lem#

$\displaystyle f(x) = \frac{3x + 2}{5x - 3}$, $\displaystyle x \ne \frac{3}{5}$. Show $(f \circ f)(x) = x$. What is $f^{-1}$?

Under­stand­ing the prob­lem#

We need $f(f(x))$: sub­sti­tute the whole frac­tion $f(x)$ in place of $x$ in $f$. If the result is $x$, then apply­ing $f$ twice brings you back, so $f$ undoes itself.

The idea#

Sim­plify the numer­a­tor and denom­i­na­tor of $f(f(x))$ sep­a­rately over the com­mon denom­i­na­tor $5x - 3$.

Step-by-step solu­tion#

Step 1. Write $f(f(x))$.

$$\displaystyle f(f(x)) = \frac{3\left(\dfrac{3x + 2}{5x - 3}\right) + 2}{5\left(\dfrac{3x + 2}{5x - 3}\right) - 3}$$

Step 2. Sim­plify the numer­a­tor.

$$\displaystyle \frac{3(3x + 2) + 2(5x - 3)}{5x - 3} = \frac{9x + 6 + 10x - 6}{5x - 3} = \frac{19x}{5x - 3}$$

Step 3. Sim­plify the denom­i­na­tor.

$$\displaystyle \frac{5(3x + 2) - 3(5x - 3)}{5x - 3} = \frac{15x + 10 - 15x + 9}{5x - 3} = \frac{19}{5x - 3}$$

Step 4. Divide; the $5x - 3$ and the $19$ can­cel.

$$\displaystyle f(f(x)) = \frac{19x}{5x - 3}\cdot\frac{5x - 3}{19} = x$$

Step 5. Since $f \circ f$ is the iden­tity, $f$ is its own inverse: $f^{-1} = f$.

Check­ing the answer#

$\displaystyle f(0) = \frac{2}{-3} = -\frac{2}{3}$; $\displaystyle f\left(-\frac{2}{3}\right) = \frac{-2 + 2}{-\frac{10}{3} - 3} = 0$. Back to $0$.

Answer#

$(f \circ f)(x) = x$, so $\displaystyle f^{-1}(x) = f(x) = \frac{3x + 2}{5x - 3}$.

Ques­tion 4: Find­ing inverses#

The prob­lem#

Find the inverse of: (a) $f(x) = 6 - 5x$; (b) $f(x) = \sqrt[3]{x - 1}$; (c) $\displaystyle f(x) = \frac{x}{x + 2}$ on $\mathbf{R} - \{-2\}$.

Under­stand­ing the prob­lem#

To find $f^{-1}$, write $y = f(x)$ and solve for $x$ in terms of $y$. The result­ing for­mula is $f^{-1}(y)$.

The idea#

Undo the oper­a­tions of $f$ in reverse order.

Step-by-step solu­tion#

Part (a)

Step 1. $\displaystyle y = 6 - 5x \;\Rightarrow\; 5x = 6 - y \;\Rightarrow\; x = \frac{6 - y}{5}$.

$$\displaystyle f^{-1}(y) = \frac{6 - y}{5}$$

Part (b)

Step 1. $y = \sqrt[3]{x - 1}$. Cube both sides: $y^3 = x - 1$.

Step 2. $x = y^3 + 1$.

$$f^{-1}(y) = y^3 + 1$$

Part (c)

Step 1. $\displaystyle y = \frac{x}{x + 2}$. Mul­ti­ply by $x + 2$: $y(x + 2) = x$, that is $xy + 2y = x$.

Step 2. Col­lect the $x$ terms: $2y = x - xy = x(1 - y)$.

Step 3. Divide by $1 - y$ (so $y \ne 1$).

$$\displaystyle f^{-1}(y) = \frac{2y}{1 - y}, \qquad y \ne 1$$

The value $y = 1$ is never taken by $f$, because $\displaystyle \frac{x}{x + 2} = 1$ would need $0 = 2$.

Check­ing the answer#

(a) $f(1) = 1$ and $\displaystyle \frac{6 - 1}{5} = 1$. (b) $f(9) = 2$ and $2^3 + 1 = 9$. (c) $\displaystyle f(2) = \frac{1}{2}$ and $\displaystyle \frac{2 \cdot \frac{1}{2}}{1 - \frac{1}{2}} = 2$.

Answer#

(a) $\displaystyle \frac{6 - y}{5}$ (b) $y^3 + 1$ (c) $\displaystyle \frac{2y}{1 - y}$, $y \ne 1$

Ques­tion 5: Is this func­tion invert­ible?#

The prob­lem#

Is $f : \{1, 2, 3\} \to \{a, b, c\}$, $f = \{(1, b), (2, a), (3, b)\}$ invert­ible? Why?

Under­stand­ing the prob­lem#

A func­tion is invert­ible exactly when it is a bijec­tion: one-one (dif­fer­ent inputs give dif­fer­ent out­puts) and onto (every ele­ment of the codomain is used).

The idea#

Check whether two inputs share an out­put.

Step-by-step solu­tion#

Step 1. $f(1) = b$ and $f(3) = b$. Two dif­fer­ent inputs give the same out­put, so $f$ is not one-one.

Step 2. (Also, $c$ is never an out­put, so $f$ is not onto either.)

Step 3. Since $f$ is not a bijec­tion, it is not invert­ible. An inverse would not know whether to send $b$ back to $1$ or to $3$.

Check­ing the answer#

Range $= \{a, b\} \ne \{a, b, c\}$, which con­firms it is not onto.

Answer#

No. $f(1) = f(3) = b$, so $f$ is not one-one (and not onto), hence not a bijec­tion.

Ques­tion 6: A self-inverse bijec­tion on $\mathbf{N}$#

The prob­lem#

Show that $f : \mathbf{N} \to \mathbf{N}$, with $f(n) = n + 1$ for odd $n$ and $f(n) = n - 1$ for even $n$, is a bijec­tion and is its own inverse.

Under­stand­ing the prob­lem#

List a few val­ues: $f(1) = 2$, $f(2) = 1$, $f(3) = 4$, $f(4) = 3$, … . The func­tion swaps each odd num­ber with the next even num­ber.

The idea#

Show $f(f(n)) = n$ for every $n$. Then $f$ has an inverse (itself), and a func­tion with an inverse is a bijec­tion.

Step-by-step solu­tion#

Step 1. Let $n$ be odd. Then $f(n) = n + 1$, which is even, so

$$f(f(n)) = (n + 1) - 1 = n$$

Step 2. Let $n$ be even. Then $f(n) = n - 1$, which is odd (and at least $1$, since $n \ge 2$), so

$$f(f(n)) = (n - 1) + 1 = n$$

Step 3. So $f \circ f = I_{\mathbf{N}}$. This says $f$ is an inverse of itself: $f^{-1} = f$.

Step 4. A func­tion is invert­ible exactly when it is a bijec­tion, so $f$ is a bijec­tion.

Check­ing the answer#

Direct check: one-one, since if $f(m) = f(n)$ then $m = f(f(m)) = f(f(n)) = n$. Onto, since any $k$ equals $f(f(k))$, the image of $f(k)$.

Answer#

$f(f(n)) = n$ for all $n$, so $f$ is a bijec­tion with $f^{-1} = f$.

Ques­tion 7: An equiv­a­lence rela­tion on $\mathbf{Z}$#

The prob­lem#

On $\mathbf{Z}$, $a \, R \, b \iff a + b$ is even. Is $R$ an equiv­a­lence rela­tion? What are its classes?

Under­stand­ing the prob­lem#

An equiv­a­lence rela­tion must be reflex­ive ($a \, R \, a$), sym­met­ric ($a \, R \, b \Rightarrow b \, R \, a$) and tran­si­tive ($a \, R \, b$ and $b \, R \, c \Rightarrow a \, R \, c$). An equiv­a­lence class is the set of all ele­ments related to a given one.

The idea#

Check the three prop­er­ties one by one using facts about even num­bers.

Step-by-step solu­tion#

Step 1. Reflex­ive: $a + a = 2a$ is even. So $a \, R \, a$ for every $a$.

Step 2. Sym­met­ric: if $a + b$ is even, then $b + a$ is the same num­ber, so it is even. So $b \, R \, a$.

Step 3. Tran­si­tive: sup­pose $a + b$ and $b + c$ are both even. Then

$$a + c = (a + b) + (b + c) - 2b$$

is a sum and dif­fer­ence of even num­bers, so it is even. Hence $a \, R \, c$.

Step 4. Classes: $a + b$ is even exactly when $a$ and $b$ are both even or both odd. So there are two classes: all even inte­gers, and all odd inte­gers.

Check­ing the answer#

$3 \, R \, 7$ ($10$ is even) and $3$ is not related to $4$ ($7$ is odd). Con­sis­tent with the two classes.

Answer#

Yes, $R$ is an equiv­a­lence rela­tion. Its classes are the even inte­gers $\{\dots, -2, 0, 2, \dots\}$ and the odd inte­gers $\{\dots, -1, 1, 3, \dots\}$.

Ques­tion 8: An equiv­a­lence rela­tion on pairs#

The prob­lem#

On $A = \mathbf{N} \times \mathbf{N}$, $(a, b) \, R \, (c, d) \iff a + d = b + c$. Show $R$ is an equiv­a­lence rela­tion.

Under­stand­ing the prob­lem#

The ele­ments are ordered pairs of nat­ural num­bers. We must prove reflex­ive, sym­met­ric and tran­si­tive prop­er­ties for all pairs.

The idea#

Each prop­erty reduces to a sim­ple equa­tion between nat­ural num­bers. For tran­si­tiv­ity, add two equa­tions and can­cel.

Step-by-step solu­tion#

Step 1. Reflex­ive: for any $(a, b)$, is $(a, b) \, R \, (a, b)$? We need $a + b = b + a$, which is true.

Step 2. Sym­met­ric: sup­pose $(a, b) \, R \, (c, d)$, so $a + d = b + c$. We need $(c, d) \, R \, (a, b)$, that is $c + b = d + a$. This is the same equa­tion read from right to left, so it holds.

Step 3. Tran­si­tive: sup­pose $(a, b) \, R \, (c, d)$ and $(c, d) \, R \, (e, f)$.

$$a + d = b + c \quad (1), \qquad c + f = d + e \quad (2)$$

Add (1) and (2):

$$a + d + c + f = b + c + d + e$$

Can­cel $c + d$ from both sides:

$$a + f = b + e$$

This says $(a, b) \, R \, (e, f)$.

Check­ing the answer#

$(3, 1) \, R \, (5, 3)$ since $3 + 3 = 1 + 5$; and $(5, 3) \, R \, (4, 2)$ since $5 + 2 = 3 + 4$; tran­si­tiv­ity pre­dicts $(3, 1) \, R \, (4, 2)$: $3 + 2 = 1 + 4$. True.

Answer#

$R$ is reflex­ive, sym­met­ric and tran­si­tive, so it is an equiv­a­lence rela­tion.

Ques­tion 9: Test­ing three prop­er­ties#

The prob­lem#

Is $R = \{(a, b) : b = a + 1\}$ on $\{1, 2, 3, 4\}$ reflex­ive, sym­met­ric, tran­si­tive?

Under­stand­ing the prob­lem#

First list $R$. Both entries must be in $\{1, 2, 3, 4\}$.

The idea#

To show a prop­erty fails, one coun­terex­am­ple is enough.

Step-by-step solu­tion#

Step 1. List $R$.

$$R = \{(1, 2), (2, 3), (3, 4)\}$$

Step 2. Reflex­ive? We would need $(1, 1) \in R$, but $1 \ne 1 + 1$. Not reflex­ive.

Step 3. Sym­met­ric? $(1, 2) \in R$ but $(2, 1) \notin R$. Not sym­met­ric.

Step 4. Tran­si­tive? $(1, 2) \in R$ and $(2, 3) \in R$, but $(1, 3) \notin R$. Not tran­si­tive.

Check­ing the answer#

In fact no pair $(a, a)$ is in $R$ and no reversed pair is in $R$, so the fail­ures are not just one-off.

Answer#

None of the three: $R$ is not reflex­ive, not sym­met­ric and not tran­si­tive.

Ques­tion 10: Ver­i­fy­ing an inverse#

The prob­lem#

Let $f(x) = 3x + 2$ and $\displaystyle g(x) = \frac{x - 2}{3}$. Ver­ify $g = f^{-1}$ by com­put­ing both com­pos­ites.

Under­stand­ing the prob­lem#

$g$ is the inverse of $f$ exactly when $g \circ f$ and $f \circ g$ are both the iden­tity, that is, both give back $x$.

The idea#

Com­pute $g(f(x))$ and $f(g(x))$.

Step-by-step solu­tion#

Step 1. $g \circ f$:

$$\displaystyle g(f(x)) = g(3x + 2) = \frac{(3x + 2) - 2}{3} = \frac{3x}{3} = x$$

Step 2. $f \circ g$:

$$\displaystyle f(g(x)) = 3\cdot\frac{x - 2}{3} + 2 = (x - 2) + 2 = x$$

Step 3. Both com­pos­ites are the iden­tity, so $g = f^{-1}$.

Check­ing the answer#

$f(1) = 5$ and $\displaystyle g(5) = \frac{3}{3} = 1$.

Answer#

$g(f(x)) = x$ and $f(g(x)) = x$, so $g = f^{-1}$.