Question

Show algebraically that f and g are inverse functions.

1.
`f(x) = -6x+3\\g(x) = (3-x)/(6)`

2.
`f(x) = -3x^(2) +5`, ≥ 0

`g(x) = \sqrt[]{(5-x)/(3) }`

Answer 1

Answer to Q1:

f⁻¹(x) = 3-x / 6 = g(x)

Explanation:

We have given two function.

f(x) = -6x+3 and g(x) = (3-x) / 6

We have to show that given two functions are inverse of each other.

Let f(x) = y

y = -6x+3

We have to separate x from above function.

Adding -3 to both sides of above equation, we have

y-3 = -6x+3-3

y-3 = -6x

Dividing by -6 to both sides of above equation , we have

(y-3) / (-6) = -6x / -6

3-y / 6 = x

Swapping above equation, we have

x = 3-y / 6

Putting x = f⁺¹(y) in above equation, we have

f⁺¹(y) = 3-y / 6

Replacing y with x, we have

f⁻¹(x) = 3-x / 6 = g(x)

Hence, f and g are inverse functions.

Answer to Q2:

f⁻¹(x) = √5-x/3 = g(x)

Explanation:

We have given two functions.

f(x) = -3x²+5

g(x) = √5-x/3

Let y = f(x)

y = -3x²+5

We have to separate x from above equation .

Adding -5 to both sides of above equation, we have

y-5 = -3x²+5-5

y-5 = -3x²

Dividing by-3 to both sides of above equation , we have

(y-5) / (-3) = -3x²/(-3)

5-y / 3 = x²

Swapping above equation, we have

x² = 5-y/3

Taking square root to both sides of above equation, we have

x = √5-y / 3

Putting x = f⁻¹(y) in above equation , we have

f⁻¹(y) = √5-y / 3

replacing y with x , we have

f⁻¹(x) = √5-x / 3 = g(x)

Hence , f and g are inverse functions.

Answer 2

Explanation:

1)

f(x) = -6x + 3

To get the inverse of this, make x be the subject of the formula.

let f(x) = y

6x = 3 - y

x = (3 - y)/6

∴ f(x) = (3 - x)/6

Truly f and g are inverse functions.

2)

f(x) = -3x²+ 5

let f(x) = y

y = -3x² + 5

3x² = 5 - y

x² = (5 - y)/3

x = √[(5 - y)/3]

f(x) = √[(5 - x)/3]

Truly f and g are inverse function.