Question

A simple cipher takes a number and codes it, using the function f(x) = 3x − 4. Find the inverse of this function, determine whether the inverse is also a function and verify the symmetrical property about the line y = x(by drawing the lines)

Answer 1

Given f(x) = 3x – 4

Let y = 3x – 4

⇒ y + 4 = 3x

⇒ x = `(y + 4)/3`

Let g(y) = `(y + 4)/3`

gof(x) = g(f(x))

= g(3x – 4)

= `(3x - 4 + 4)/3`

= `(3x)/3`

gof(x) = x

and fog(y) = f(g(y))

= `f((y + 4)/3)`

= `3((y + 4)/3)`

= y + 4 – 4 = y

fog(y) = y

Hence g of = I_x and fog = I_y

This shows that f and g are bijections and inverses of each other.
Hence f is bijection and f^–1(y) = `(y + 4)/3`

Replacing y by x we get f^–1(x) = `(x + 4)/3`

The line y = x

x	0	1	– 1	2	– 2	3	– 3
y	0	1	– 1	2	– 2	3	– 3

f(x) = The line y = 3x – 4

When x = 0 ⇒ y = 3 × 0 – 4 = – 4

When x = 1 ⇒ y = 3 × 1 – 4 = – 1

When x = – 1 ⇒ y = 3 × – 1 – 4 = – 7

When x = 2 ⇒ y = 3 × 2 – 4 = 2

When x = – 2 ⇒ y = 3 × – 2 – 4 = – 10

When x = 3 ⇒ y = 3 × 3 – 4 = 5

When x = – 3 ⇒ y = 3 × – 3 – 4 = – 13

x	0	1	– 1	2	– 2	3	– 3
y	– 4	– 1	– 7	2	– 10	5	– 13

The line y = `(x + 4)/3`

When x = 0 ⇒ y = `(0 + 4)/3 = 4/3`

 When x = 1 ⇒ y = `(1 + 4)/3 = 5/3`

When x = – 1 ⇒ y = `(-1 + 4)/3` = 1

When x = 2 ⇒ y = `(2 + 4)/3` = 2

When x = – 2 ⇒ y = `(- 2 + 4)/3 = 2/3`

When x = 3 ⇒ y = `(3 + 4)/3 = 7/3`

When x = – 3 ⇒ y = `(- 3 + 4)/3 = 1/3`

x	0	1	– 1	2	– 2	3	– 3
y	`4/5`	`5/3`	1	2	`2/3`	`7/3`	`1/3`

From te graph, the lines y = 3x – 4 and y = `(x+ 4)/3` are symmertical about the line y = x.