Rational & exp functions 1 · The reciprocal function y = 1/x

Module 15 of 46Lesson 1 of 8 · Rational & exp functions 1 · The reciprocal function y = 1/x

Rational, exponential & logarithmic functions; solving equations · Essential understanding 1

The reciprocal function y = $\dfrac{1}{x}$

IB AA SL 2.8with Prof. Kitso

💻 Using the TI-Nspire CX Student Software on a laptop

The screens in these notes are drawn exactly as the Student Software shows them. The keypad on the left of the software works like the handheld — click its keys with the mouse.
menu: click menu on the on-screen keypad (or Documents Toolbox ▸ 🔧). The menu numbers and letters are the same as in these notes.
enter = Enter · tab = Tab (next box) · esc = Esc · del = Backspace.
Powers: type ^ (Shift+6), the power, then press → to come back down before typing more. x² = ^2 then →.
Fraction: type / · √: type sqrt( · π: type pi · eˣ: type exp( · store: type :=. For EE (×10ⁿ) and i, click them on the keypad.
New page: Ctrl+I (or the Insert menu) · change page: Ctrl+← / Ctrl+→ · graph entry line: Ctrl+G · var: Ctrl+L.
Radian / Degree: click Settings in the status bar at the bottom of the window. The mode (RAD or DEG) shows in the top bar.
IB exam mode (Press-to-Test): everything in these notes works on the non-CAS TI-Nspire with the IB exam settings. In exam mode there are no dot-product, cross-product, unit-vector or eigenvalue keys and no Diff Eq graphs — the notes show the allowed way (type the components, use polyRoots on the characteristic equation, do Euler steps on a Calculator page).
If you type nDeriv(, the software shows it as centralDiff( — it is the same command.

The big idea: The reciprocal function

f(x) = \dfrac{1}{x}

turns every number into "one divided by it". Its graph is a curve in two pieces (a hyperbola) that gets closer and closer to the two axes but never touches them. Those never-touched lines are called asymptotes.

By the end of this lesson you can:

sketch $y = \dfrac{1}{x}$ and explain its shape from a table of values
write down its asymptotes, domain and range
show that $f(x) = \dfrac{1}{x}$ is self-inverse and explain what that means for the graph
sketch $y = \dfrac{a}{x-h} + k$ and find its asymptotes and intercepts
use a reciprocal model in a context (sharing a cost)

🧑🏾 Imagine this…

Neo shares a P600 taxi fare with some friends. Alone she pays P600. With one friend, each pays P300. With 5 friends in total, each pays P120. The more people, the smaller each share:

\dfrac{600}{n}

. Each share gets tiny but never reaches zero, however many people come. That "gets close but never touches" behaviour is exactly what the graph of

\dfrac{1}{x}

shows.

Reciprocal

the reciprocal of

x

\dfrac{1}{x}

. The reciprocal of 4 is

\dfrac{1}{4}

; the reciprocal of

\dfrac{2}{3}

\dfrac{3}{2}

. Zero has no reciprocal.

Asymptote

a line that a curve gets closer and closer to, without ever touching it (as

x

y

gets very large or very close to a number).

Vertical asymptote (VA)

a vertical line

x = c

. The curve shoots up or down as

x

approaches

c

Horizontal asymptote (HA)

a horizontal line

y = c

. The curve levels off towards it as

x

becomes very large (positive or negative).

Self-inverse

a function that is its own inverse:

f^{-1}(x) = f(x)

, or

f(f(x)) = x

★ The graph of y =

\dfrac{1}{x}

Domain: all real $x$ except $x = 0$ , written $x \neq 0$ . Range: all real $y$ except $y = 0$ .
Asymptotes: the $y$ -axis $x = 0$ (vertical) and the $x$ -axis $y = 0$ (horizontal).
Two branches: one in the top right ( $x > 0, y > 0$ ) and one in the bottom left ( $x < 0, y < 0$ ). There are no intercepts: $\dfrac{1}{x}$ is never 0, and $x = 0$ is not allowed.
Symmetry: the graph is symmetric about the line $y = x$ (and also about $y = -x$ ). This is why $f$ is its own inverse.
Self-inverse: $f(f(x)) = \dfrac{1}{\frac{1}{x}} = x$ , so $f^{-1}(x) = \dfrac{1}{x}$ . The same is true for any $\dfrac{a}{x}$ .

★ The family y =

\dfrac{a}{x - h}

+ k

Start from $y = \dfrac{1}{x}$ . The graph of $y = \dfrac{a}{x-h} + k$ is a stretch (factor $a$ , with a reflection in the $x$ -axis if $a < 0$ ), then a shift right $h$ and up $k$ .
Vertical asymptote: $x = h$ (the value that makes the denominator 0). Horizontal asymptote: $y = k$ .
Domain: $x \neq h$ . Range: $y \neq k$ .
Intercepts: put $x = 0$ for the $y$ -intercept; put $y = 0$ and solve for the $x$ -intercept.
If $a > 0$ the branches lie top-right and bottom-left of the centre $(h, k)$ ; if $a < 0$ they lie top-left and bottom-right.

⚠️ Watch out

Do not mix up the reciprocal $\dfrac{1}{f(x)}$ with the inverse $f^{-1}(x)$ . They are different things. Only for $f(x) = \dfrac{1}{x}$ do the two happen to agree. (Note $\dfrac{1}{x} = x^{-1}$ is a power, while $f^{-1}$ is a function inverse.)
An asymptote is a dashed line in a sketch, and you must label its equation (for example $x = 3$ ). The curve must never cross a vertical asymptote and must visibly bend towards each asymptote.
In $y = \dfrac{a}{x-h}+k$ , a minus in the bracket moves the graph right: $\dfrac{1}{x-3}$ has its asymptote at $x = +3$ , and $\dfrac{1}{x+2}$ at $x = -2$ .

🧪 Try it yourselfSlide the hyperbola about

This is

y = \dfrac{a}{x-h} + k

. Start with a = 1, h = 0, k = 0 (the basic

\dfrac{1}{x}

). Then try h = 3, k = 1, a = 2: you get Example 1.3. Notice the vertical asymptote follows h and the horizontal one follows k. Make a negative and watch the branches swap sides.

Worked examples

Example 1.1Meet y =

\dfrac{1}{x}

Let

f(x) = \dfrac{1}{x}

(a) Complete a table of values for $x = -4, -2, -1, -0.5, 0.5, 1, 2, 4$ .
(b) Describe what happens to $y$ when $x$ is very close to 0 on the right, and when $x$ is very large.
(c) Write down the equations of the asymptotes and the range.

🧠 Think it through

Work out $1 \div x$ for each value. Negative in, negative out.
Try bigger tests: $x = 0.01$ gives 100; $x = 100$ gives 0.01.
Asymptotes come from the two things that can never happen: $x = 0$ and $y = 0$ .

✍️ By hand

(a)

f(-4) = -\dfrac{1}{4},\; f(-2) = -\dfrac{1}{2},\; f(-1) = -1,\; f(-0.5) = -2

f(0.5) = 2,\; f(1) = 1,\; f(2) = \dfrac{1}{2},\; f(4) = \dfrac{1}{4}

(b)

x = 0.1 \to y = 10;\; x = 0.01 \to y = 100;\; x = 0.001 \to y = 1000

. So

y

grows without limit as

x \to 0^+

. For large

x

y \to 0

from above.

x = 0

; horizontal asymptote

y = 0

; range

y \neq 0

🔢 On your TI-Nspire CX II

Open Lists & Spreadsheet. Name column A x and type the 8 values. Name column B y, then in the formula row (the grey cell under the name) type =1/x and press enter ↵💻 Enter.
The column fills itself. Use ctrl💻 Ctrl doc ▾💻 File menu ▸ Graphs to see the curve in two pieces.

✅ Answer (a) values as above. (b) As x approaches 0 from the right, y grows without limit; for large x, y gets close to 0. (c) Asymptotes x = 0 and y = 0; range y ≠ 0.

Example 1.2A self-inverse function

Let

f(x) = \dfrac{1}{x}

x \neq 0

, and

g(x) = \dfrac{5}{x}

x \neq 0

(a) Find $f(f(x))$ .
(b) Find the inverse function $g^{-1}(x)$ .
(c) What does this tell you about the graph of $g$ ?

🧠 Think it through

(a) Put $\dfrac{1}{x}$ inside $f$ in place of $x$ .
(b) Swap $x$ and $y$ in $y = \dfrac{5}{x}$ , then make $y$ the subject.
(c) A function equal to its own inverse is symmetric when reflected in the line $y = x$ .

✍️ By hand

(a)

f(f(x)) = f\!\left(\dfrac{1}{x}\right) = \dfrac{1}{\;\frac{1}{x}\;} = x

(M1)(A1)

(b)

y = \dfrac{5}{x}\;\Rightarrow\; x = \dfrac{5}{y}\;\Rightarrow\; y = \dfrac{5}{x}

(M1) so

g^{-1}(x) = \dfrac{5}{x} = g(x)

(A1)

(c)

g

is self-inverse, so its graph is unchanged when reflected in

y = x

: it is symmetric about the line

y = x

✅ Answer (a) x. (b) g⁻¹(x) =

\dfrac{5}{x}

, x ≠ 0. (c) g is self-inverse, so its graph is symmetric about y = x.

Check on the GDC: g(2) = 2.5 and g(2.5) = 2. Going 2 → 2.5 → 2 shows g undoes itself.

Example 1.3Sketch y =

\dfrac{a}{x - h}

+ k

Sketch

y = \dfrac{2}{x-3} + 1

. State the asymptotes, intercepts, domain and range.

🧠 Think it through

Compare with $\dfrac{a}{x-h}+k$ : $a = 2,\; h = 3,\; k = 1$ .
VA is $x = 3$ , HA is $y = 1$ . Since $a > 0$ the branches are top-right and bottom-left of the centre $(3, 1)$ .
Intercepts: put $x = 0$ , then put $y = 0$ .

✍️ By hand

Asymptotes:

x = 3

and

y = 1

. (A1)(A1)

y

-intercept:

x = 0 \Rightarrow y = \dfrac{2}{-3} + 1 = \dfrac{1}{3}

(A1)

x

-intercept:

\dfrac{2}{x-3} + 1 = 0 \Rightarrow \dfrac{2}{x-3} = -1 \Rightarrow 2 = -(x-3) \Rightarrow x = 1

(A1)

Domain

x \neq 3

; range

y \neq 1

🔢 On your TI-Nspire CX II

Enter 2/(x-3)+1 as f1(x). Use Analyze Graph ▸ Zero to confirm the $x$ -intercept at (1, 0).

✅ Answer VA x = 3, HA y = 1; y-intercept (0,

\dfrac{1}{3}

), x-intercept (1, 0); domain x ≠ 3, range y ≠ 1.

In your own sketch draw the two asymptotes dashed, label their equations, mark both intercepts, and keep each branch on the correct side of the centre (3, 1).

Example 1.4A negative a: reading the transformations

The graph of

y = -\dfrac{4}{x+2} - 3

is obtained from

y = \dfrac{1}{x}

by a sequence of transformations.

(a) Describe the transformations.
(b) Write down the asymptotes and find both intercepts.
(c) State which side of the asymptotes each branch lies on.

🧠 Think it through

Write it as $\dfrac{a}{x-h}+k$ : here $a = -4,\; h = -2,\; k = -3$ .
Negative $a$ means a reflection in the $x$ -axis as well as the stretch.
Use a test point (the $y$ -intercept) to see which branch it belongs to.

✍️ By hand

(a) Vertical stretch, scale factor 4; reflection in the

x

-axis; translation

\begin{pmatrix}-2\\-3\end{pmatrix}

(A1)(A1)(A1)

(b) VA

x = -2

, HA

y = -3

y

-intercept:

x = 0 \Rightarrow y = -\dfrac{4}{2} - 3 = -5

x

-intercept:

-\dfrac{4}{x+2} = 3 \Rightarrow x + 2 = -\dfrac{4}{3} \Rightarrow x = -\dfrac{10}{3}

(c)

a < 0

: the branches are top-left and bottom-right of

(-2,-3)

. Check:

(0,-5)

is to the right of

x=-2

and below

y=-3

\checkmark

🔢 On your TI-Nspire CX II

Plot to check the shape and the intercepts.

✅ Answer (a) stretch ×4 (vertical), reflection in the x-axis, translation (−2, −3). (b) x = −2, y = −3; y-intercept (0, −5), x-intercept (−

\dfrac{10}{3}

, 0). (c) top-left and bottom-right.

Example 1.5Sharing a cost

A school hires a bus for a trip to Kasane for P2400. Each of the

n

learners pays an equal share of the bus plus a P85 entrance fee, so the cost per learner is

c = \dfrac{2400}{n} + 85

pula.

(a) Find the cost per learner when 20 learners go.
(b) Find the smallest number of learners so that the cost per learner is at most P150.
(c) Write down the horizontal asymptote and explain what it means in this context.

🧠 Think it through

(a) Substitute $n = 20$ .
(b) Solve $\dfrac{2400}{n} + 85 \le 150$ . Subtract 85, then rearrange (n is positive, so the inequality does not flip).
(c) The model is $\dfrac{a}{n}+k$ with $k = 85$ .

✍️ By hand

(a)

c = \dfrac{2400}{20} + 85 = 120 + 85 = 205

(b)

\dfrac{2400}{n} \le 65 \;\Rightarrow\; 2400 \le 65n \;\Rightarrow\; n \ge \dfrac{2400}{65} = 36.9\ldots

n = 37

c = 85

. However many learners go, the cost per learner gets close to P85 (the entrance fee) but never falls below it.

🔢 On your TI-Nspire CX II

Check by substituting: 36 learners is too few, 37 is enough.

✅ Answer (a) P205. (b) 37 learners. (c) c = 85: the cost per learner approaches P85 but never goes below it.

Challenge questions

⭐ Challenge 1.6Inverse of a reciprocal-type functionPaper 1 style

Let

g(x) = \dfrac{6}{x+3} + 2

x \neq -3

(a) Write down the equations of the asymptotes.
(b) Find the $x$ - and $y$ -intercepts.
(c) Find $g^{-1}(x)$ and state its domain.

✏️ Try it on paper and on your calculator first. Then check.

Show the full solution

🧠 Think it through

Asymptotes: bracket value $-3$ , vertical shift $+2$ .
Intercepts: sub $x = 0$ , then solve $g(x) = 0$ .
Inverse: write $y = \dfrac{6}{x+3}+2$ , swap, rearrange. The domain of $g^{-1}$ is the range of $g$ .

✍️ By hand

(a) VA

x = -3

; HA

y = 2

(b)

x = 0

y = \dfrac{6}{3} + 2 = 4

y = 0

\dfrac{6}{x+3} = -2 \Rightarrow x + 3 = -3 \Rightarrow x = -6

(c)

x = \dfrac{6}{y+3}+2 \Rightarrow x - 2 = \dfrac{6}{y+3} \Rightarrow y + 3 = \dfrac{6}{x-2} \Rightarrow y = \dfrac{6}{x-2} - 3

(M1)(A1)

g^{-1}(x) = \dfrac{6}{x-2} - 3

x \neq 2

(the range of

g

y \neq 2

). (A1)

✅ Answer (a) x = −3, y = 2. (b) (0, 4) and (−6, 0). (c) g⁻¹(x) =

\dfrac{6}{x - 2}

− 3, x ≠ 2.

⭐ Challenge 1.7A hyperbola meets a linePaper 2 style

Let

f(x) = \dfrac{4}{x-2} + 3

x \neq 2

, and

h(x) = x + 1

(a) Write down the asymptotes of the graph of $f$ and find its $x$ -intercept.
(b) Show that $f(x) = h(x)$ leads to $x^2 - 4x = 0$ .
(c) Hence find the coordinates of the points where the two graphs meet.
(d) Use your GDC to check, and say how many times the graphs meet.

✏️ Try it on paper and on your calculator first. Then check.

Show the full solution

🧠 Think it through

(a) Read the asymptotes off. The $x$ -intercept comes from $f(x) = 0$ .
(b) Multiply both sides by $(x-2)$ and collect terms.
(c) Solve the quadratic, then find the $y$ -values using $h$ .
(d) Graph both and use Intersection.

✍️ By hand

(a) VA

x = 2

, HA

y = 3

\dfrac{4}{x-2} = -3 \Rightarrow x - 2 = -\dfrac{4}{3} \Rightarrow x = \dfrac{2}{3}

(b)

\dfrac{4}{x-2} + 3 = x + 1 \Rightarrow 4 + 3(x-2) = (x+1)(x-2)

(M1)

3x - 2 = x^2 - x - 2 \Rightarrow x^2 - 4x = 0

(AG)

(c)

x(x-4) = 0 \Rightarrow x = 0

x = 4

h(0) = 1

h(4) = 5

. Points

(0, 1)

and

(4, 5)

. (A1)(A1)

🔢 On your TI-Nspire CX II

Graph 4/(x-2)+3 and x+1. Menu ▸ Analyze Graph ▸ Intersection. The two points are (0, 1) and (4, 5).

✅ Answer (a) x = 2, y = 3; x-intercept (

\dfrac{2}{3}

, 0). (b) shown. (c) (0, 1) and (4, 5). (d) They meet twice.

Next: Lesson 2 · Rational functions

\dfrac{ax + b}{cx + d}

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL Rational & exp functions 1 · The reciprocal function y = 1/x

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL

The reciprocal function y = $\dfrac{1}{x}$

Worked examples

Challenge questions