# Inverse Functions

## Definition

The inverse is shown by putting a little “-1” after the function name, like this:

f-1(y), where y = f(x)

It says “f inverse of y

Here we have the function f(x) = 3x+2

The flow is

Inverse Functions make 3x + 2 back to x

, substitute y = 3x + 2,

then

So, the inverse of f(x) = 3x + 2 is f-1(y) =

Let’s put x = 3 on f(x):

f(3) = 3×3 + 2 = 11

Then put y = 11 on f-1(y):

f-1(11) = = 3

So,

f-1(y)=f-1(f(x)) = x

Then, how about f (f-1(x)) ?

Because f-1(y)= x, then f (f-1(y)) = f (x) = y

f (f-1(y)) = y. Which means f (f-1(x)) = x

So

f-1(f(x)) = f (f-1(x)) = x

## Inverses of Common Functions

So far is easy, because we know the inverse of Add is Subtract, and the inverse of Multiply is Divide. Ho about other functions?

## Invers Function Examples

1. Determine the invers function

• f(x) = 2x – 3
• a(x) = ½ x – 4
• f(y) = (y-2)3 – 2
• h(x) = (-x + 3) / 5
• g(x) = 2sinx + 1
• f(x) = 2x – 3

replace f(x) with y

y = 2x – 3
y + 3 = 2x
(y+3)/2 = x

Then f-1(x) = (x+3) / 2

• a(x) = ½ x – 4

replace a(x) with y

y = ½ x – 4
y + 4 = ½ x
2y + 8 = x

Then a-1(x) = 2x + 8

• f(y) = (y-2)3 – 2

replace f(y) with a

a = (y-2)3 – 2
a + 2 = (y-2)3
3√(a+2) = y – 2
3√(a+2) + 2 = y

Then f-1(y) = 3√(y+2) + 2

• h(x) = (-x + 3) / 5

replace h(x) with y

y = (-x + 3) / 5
5y = -x + 3
5y – 3 = -x
x = -5y + 3

then h-1(x) = -5x + 3

• g(x) = 2sinx + 1

replace g(x) with y

y = 2 sinx + 1
y – 1 = 2 sinx
(y – 1) / 2 = sin x
x = arc sin{(y – 1) / 2}
then g-1(x) = arc sin{(x – 1) / 2}

2. Determine the domain of invers function

• g(x) = -¼x – ¾
• k(x) = 4x / (2x-1)
• f(x) = {√(2x-1)} / 4
• b(x) = 2x¾ + 1
•       g(x) = -¼x – ¾
y = -¼x – ¾
y + ¾ = -¼x
-4y – 3 = x
g-1(x) = -4x – 3

then D(g-1) = all real numbers.

•           k(x) = 4x / (2x-1)
y = 4x / (2x-1)
2xy – y = 4x
2xy – 4x = y
x(2y – 4) = y
x = y / (2y – 4)
k-1(x) = x / (2x – 4)

then D(k-1) = all real numbers except 2.

•                   f(x) = {√(2x-1)} / 4
y = {√(2x-1)} / 4
4y  = √(2x-1)
(4y)2 = 2x – 1
16y2 = 2x – 1
16y2 + 1 = 2x
(16y2 + 1) / 2 = x
f-1(x) = (16x2 + 1) / 2

then D(f-1) = all real numbers

•              b(x) = 2x¾ + 1
y = 2x¾ + 1
y – 1 = 2x¾
(y – 1) / 2 = x¾
¾ √{(y-1)/2} = x
b-1(x) = ¾ √{(x-1)/2}

then D(b-1) = all real numbers x ≠ 1

3. If f(x) = 2x / (x+1) then f-1(3) = …

f(x) = 2x / (x+1)
y = 2x / (x+1)
xy + y = 2x
xy – 2x = -y
x(y-2) = -y
x = -y / (y-2)
f-1(x) = -x / (x-2)

then f-1(3) = -3 / (3-2) = -3

4. If g(x) = 2x – 1, h(x) = x2 + x – 1 then g-1(1) + g(0) – h(2) = …

Determine g-1(-1)

g(x) = 2x – 1
y = 2x – 1
y + 1 = 2x
x = (y+1) / 2
g-1(x) = (x+1) / 2
g-1(-1) = 0

Then

g-1(1) + g(0) – h(2)
0 + {2(0) – 1} – {22 + 2 – 1}  = -6

5. If f-1(x) = x2 – 2x + 1 then f(1) = …

Remember concept {f-1(x)}-1 = f(x)

Then

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

replace f-1(x) with y

y = x2 – 2x + 1
y = (x – 1)2
√y = x – 1
√y + 1 = x
f(x) = √x + 1

so, f(1) = √1 + 1 = 2