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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)
Solution
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 = Ix and fog = Iy
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.
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