Deriving the Inverse of a Function (Simple Functions)

1. Deriving the Inverse of a
Function
Presented by:
ALONA HALL

2. Objectives of the Lesson
• Explain the concept of an ‘inverse function’ by using a mapping
diagram
• Demonstrate how the inverse of a function is found.

3. The Concept
The function f (x) = 2x +5 maps the domain elements to the range
elements in the diagram shown.
i.e. 𝑓 1 = 2 1 + 5 = 7
You substitute the domain (input) elements into the function in order
to obtain the range (output) elements.
If, however, we were to map the range elements to the domain
elements then we would need another function that reverses the first
function. That function is known as the inverse function.

4. 𝑓−1
𝑥 =
𝑥 − 5
2
e.g. 𝑓−1 7 =
7 −5
2
=
2
2
= 1
In the inverse function, the range elements become the
domain elements and vice versa.
Such a fact is very important when determining the inverse
function of a given function

5. Example
Determine the inverse of the following functions:
a) f(x) = 3x – 5
b) g(x) =
𝑥
7
+ 4
c) h(x)=
5𝑥+4
𝑥+9

6. Solutions
The rules used for transposition will become very useful for this topic.
a) f(x) = 3x – 5
First: rewrite the function as y = 3x – 5
Then: interchange x for y and y for x (x y), i.e. x = 3y - 5
Finally: transpose for y (make y the subject)
y was multiplied by 3, then 5 was subtracted
Reversing the process we get: add 5 but show it on the other side of function
x + 5 = 3y
Now divide by 3 (show it on the opposite side):
𝑥+5
3
= 𝑦
Rewrite the inverse function using the notation: 𝑓−1 𝑥 =
𝑥+5
3

7. Solutions (cont’d)
b) g(x) =
𝑥
7
+ 4
• 𝑦 =
𝑥
7
+ 4
• 𝑥 ↔ 𝑦 ∶ 𝑥 =
𝑦
7
+ 4
• 𝑡𝑟𝑎𝑛𝑠𝑝𝑜𝑠𝑒 𝑓𝑜𝑟 𝑦:
𝑦 𝑤𝑎𝑠 𝑑𝑖𝑣𝑖𝑑𝑒𝑑 𝑏𝑦 7 𝑡ℎ𝑒𝑛 4 𝑎𝑑𝑑𝑒𝑑 … 𝑠𝑜 𝑎𝑠𝑢𝑏𝑡𝑟𝑎𝑐𝑡 4 𝑡ℎ𝑒𝑛 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑏𝑦 7
𝑥 − 4 =
𝑦
7
7 𝑥 − 4 = 𝑦
𝑔−1 𝑥 = 7(𝑥 − 4)

8. Solutions (cont’d)
c) h(x)=
5𝑥+4
𝑥+9
• 𝑦 =
5𝑥+4
𝑥+9
• 𝑥 ↔ 𝑦: 𝑥 =
5𝑦+4
𝑦+9
• Transpose for y:
𝑅𝑒𝑚𝑜𝑣𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛: 𝑦 + 9 𝑖𝑠 𝑑𝑖𝑣𝑖𝑑𝑖𝑛𝑔 𝑜𝑛 𝑡ℎ𝑒 𝑟𝑖𝑔ℎ𝑡
ℎ𝑎𝑛d 𝑠ide RHS so 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑏𝑦 𝑦 + 9 𝑜𝑛
𝑙𝑒𝑓𝑡 ℎ𝑎𝑛𝑑 𝑠𝑖𝑑𝑒(𝐿𝐻𝑆)
𝑖𝑛𝑡𝑟𝑜𝑑𝑢𝑐𝑒 𝑏𝑟𝑎𝑐𝑘𝑒𝑡𝑠
𝑥 𝑦 + 9 = 5𝑦 + 4
𝑥𝑦 + 9𝑥 = 5𝑦 + 4
𝐺𝑟𝑜𝑢𝑝 𝑦 𝑡𝑒𝑟𝑚𝑠 𝑜𝑛 𝑜𝑛𝑒 𝑠𝑖𝑑𝑒 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑛𝑜𝑛
− 𝑦 𝑡𝑒𝑟𝑚𝑠 𝑜𝑛 𝑡ℎ𝑒 𝑜𝑡ℎ𝑒𝑟:
𝑥𝑦 − 5𝑦 = 4 − 9𝑥
𝐹𝑎𝑐𝑡𝑜𝑟 𝑜𝑢𝑡 𝑦:
𝑦 𝑥 − 5 = 4 − 9𝑥
𝑥 − 5 𝑖𝑠 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦𝑖𝑛𝑔 𝑜𝑛 𝑡ℎ𝑒 𝐿𝐻𝑆, 𝑡ℎ𝑢𝑠 𝑖𝑓
𝑡𝑟𝑎𝑛𝑠𝑝𝑜𝑠𝑖𝑛𝑔 𝑥 − 5 𝑠ℎ𝑜𝑢𝑙𝑑 𝑑𝑖𝑣𝑖𝑑𝑒 𝑜𝑛 𝑡ℎ𝑒 𝑅𝐻𝑆
𝑦 =
4 − 9𝑥
𝑥 − 5
ℎ−1
𝑥 =
4 − 9𝑥
𝑥 − 5