Higher Mathematics · Classes 9-10 · Chapter 1

One-One, Onto and Inverse Functions.

A function means every member of the domain fires exactly one arrow. Now ask from the other side: looking at the codomain, did anybody catch two arrows, and did anybody catch none? Those two answers decide whether you can turn the arrows round and still have a function.

1. The book's three arrow diagrams

In the first diagram, from A = {a, b, c} to B = {1, 2, 3, 4}, a → 1, b → 2 and c → 4. No two arrows land on the same place, so the function is one-one. But no arrow reaches 3, so 3 has no pre-image, and the function is not onto.

In the middle diagram, from A = {a, b, c} to C = {1, 2}, a → 1, b → 2 and c → 2. Everyone in C catches an arrow, so the function is onto. But 2 is the image of both b and c, so it is not one-one.

In the right-hand diagram, from A = {a, b, c} to D = {1, 2, 3}, a → 2, b → 1 and c → 3. This function is both one-one and onto. So turning the arrows round gives a function from D to A, called the inverse function: f⁻¹(1) = b, f⁻¹(2) = a, f⁻¹(3) = c. The first two diagrams cannot do this: in one, 3 has nowhere to go; in the other, 2 would have to go two ways.

Now the arrows are yours. Pick an arrow, then say where it goes. Drop 4 from B or put it back. Watch when one-one and onto happen together, and what turning the arrows round does then.

Now name them yourself. Six arrow diagrams, one at a time.

2. One-one functions

Definition 10: f is a one-one function if different members of its domain always have different images: x₁, x₂ ∈ dom f and x₁ ≠ x₂ imply f(x₁) ≠ f(x₂). The same thing said the other way round: f(x₁) = f(x₂) implies x₁ = x₂.

For a proof the second form is easier: assume f(a) = f(b) and show a = b. Example 30: for f(x) = 3x + 5, 3a + 5 = 3b + 5 ⟹ 3a = 3b ⟹ a = b, so f is one-one. In the same way 2x − 3, 5x − 3 and x³ are one-one.

To show a function is not one-one, a single example is enough. Example 31: for F(x) = x², x₁ = −1 and x₂ = 1 are different but F(−1) = F(1) = 1, so F is not one-one. For the same reason x² + 1, (x − 1)² and |x| are not one-one.

A fraction needs one trick. Example 36: for f(x) = (2x − 1)/(2x + 3), set f(x₁) = f(x₂) and subtract 1 from both sides: −4/(2x₁ + 3) = −4/(2x₂ + 3), so 2x₁ + 3 = 2x₂ + 3 and x₁ = x₂. f is one-one.

3. Onto functions

Definition 11: f : A → B is onto if for every b ∈ B there is an a ∈ A with f(a) = b; that is, B = range f. For example, A = {1, 2, 3}, B = {5, 7, 9} with 1 → 5, 2 → 7, 3 → 9.

The test: solve y = f(x) for x, and see whether every y gets an x in the domain. For f(x) = x + 5 (R → R), x = y − 5, which is always real, so f is onto. But f : {1, 2, 3, 4} → R, f(x) = 2x + 1 has range only {3, 5, 7, 9}, which is not R: one-one, yet not onto.

One y with no pre-image is enough to fail. For f(x) = (x − 3)/(x + 5) (codomain R), y = (x − 3)/(x + 5) gives x = (3 + 5y)/(1 − y); at y = 1 the denominator is zero and there is no x. The range is R − {1}, so f is not onto. Likewise (2x + 2)/(x − 1) has range R − {2}.

The book's activity: f : {−2, −1, 0, 1, 2} → {−8, −1, 0, 1, 8}, f(x) = x³. Five different x give five different images, and every member of the codomain is somebody's image. So it is one-one and onto.

4. The inverse function

Definition 9: if f : A → B is one-one and onto, a function g : B → A is described by g(b) = a if and only if f(a) = b, for every b ∈ B. This g is the inverse function of f, written f⁻¹.

Turning round the arrows of any function gives a relation, but it may not be a function. If f is not one-one, some y has two pre-images, and the reversed arrows send y two ways: then f⁻¹ is not a function.

Finding the formula takes three moves: write y = f(x), express x in terms of y, then put x where y was. Example 33: y = 3x + 1 ⟹ x = (1/3)(y − 1), so f⁻¹(x) = (1/3)(x − 1).

In the same example the domain is 0 ≤ x ≤ 2. The end points are (0, 1) and (2, 7), so range f = {y : 1 ≤ y ≤ 7} = [1, 7]. f is one-one, and the graph of f⁻¹ is y = (1/3)(x − 1), 1 ≤ x ≤ 7: the range of f is the domain of f⁻¹.

Example 33's f(x) = 3x + 1, 0 ≤ x ≤ 2, and its inverse on the same graph paper. Slide x: one dot on f, and its partner on f⁻¹.

Example 34: f(x) = x + 5 and g(x) = x − 5 (R → R). f is one-one and onto. If f⁻¹(x) = y then f(y) = x, so y + 5 = x and y = x − 5 = g(x). The domains are the same, so f⁻¹ = g.

In general f(x) = ax + b (a, b ∈ R, a ≠ 0, R → R) is one-one and onto, with f⁻¹(x) = (x − b)/a.

Powers follow the same moves. F(x) = x³ (R → R) is one-one and onto, with F⁻¹(x) = x^(1/3) = ∛x. For f(x) = x³ + 5, f⁻¹(x) = (x − 5)^(1/3); for x³ + 3, F⁻¹(3) = 0. And from straight lines: 5x − 4 ⟹ (x + 4)/5; for x + 1, f⁻¹(2) = 1; the inverse of 3x + 3 is (x − 3)/3.

For the shape (ax + b)/(cx + d), clear the denominator first. The book's activity: f(x) = (4x − 9)/(x − 2) ⟹ xy − 2y = 4x − 9 ⟹ f⁻¹(x) = (2x − 9)/(x − 4), x ≠ 4. Then f⁻¹(1) = 7/3, f⁻¹(−1) = 11/5, f⁻¹(−3) = 15/7. And 4f⁻¹(x) = x gives x² − 12x + 36 = 0, so x = 6.

More from the activity: 3/(x − 1) ⟹ f⁻¹(x) = (x + 3)/x, f⁻¹(3) = 2. 2x/(x − 2) ⟹ f⁻¹(x) = 2x/(x − 2): the function is its own inverse. (2x + 3)/(2x − 1) ⟹ (x + 3)/(2(x − 1)). (2x + 2)/(x − 1) ⟹ (x + 2)/(x − 2), f⁻¹(3) = 5, and f⁻¹(p) = kp gives k = (p + 2)/(p(p − 2)).

The end of Example 36: for f(x) = (2x − 1)/(2x + 3), y(2x + 3) = 2x − 1 ⟹ x = (1 + 3y)/(2(1 − y)), so f⁻¹(x) = (1 + 3x)/(2(1 − x)). If 2f⁻¹(x) = x, then (1 + 3x)/(1 − x) = x ⟹ x² + 2x + 1 = 0 ⟹ x = −1.

For a square root, square both sides to free x. y = √(x − 4) ⟹ x = y² + 4, so f⁻¹(x) = x² + 4 (x ≥ 0). y = √(1 − 2x) ⟹ F⁻¹(x) = (1 − x²)/2. y = √(2 − 5x) ⟹ f⁻¹(x) = (2 − x²)/5, f⁻¹(2) = −2/5.

Pick a function, then find its inverse one move at a time. Do not forget to swap the letter at the end.

5. f⁻¹(k) without the whole formula

If only one value is asked for, you do not need the whole formula. Let a = f⁻¹(k); then f(a) = k, and you solve that equation. Example 32: for f(x) = x/(x − 2), a/(a − 2) = 2 ⟹ a = 2a − 4 ⟹ a = 4, so f⁻¹(2) = 4. Going forwards, f(5) = 5/3 = 1⅔.

Give a function and k. For the second kind of question, c·f⁻¹(x) = x, choose it and give c.

Last, x². F(x) = x² (R → R) is not one-one, so F⁻¹ is not a function. Its range is {y ∈ R : y ≥ 0}, because no real square is negative. Read F⁻¹ as a relation and F⁻¹(x) = ±√x: for x² + 1, the pre-images of 5 are {2, −2}. But take both domain and codomain to be R₊ and F : R₊ → R₊, F(x) = x² is one-one and onto, with range R₊ and F⁻¹(x) = √x.

6. Check yourself

The first is from a definition.

For the second, solve y = x + 2 for x.

The last one has two answers to find.

The whole note on eight cards.

Source: Higher Mathematics, Classes 9-10 (NCTB), Chapter 1 “Set and Function”, the “One-One, Onto and Inverse Functions” part.