inverse of f(x)=sqrt(4-x)+3 — step-by-step solution

The function is strictly monotone on

x \leq 4

: it has an inverse on its range

y \geq 3

Isolate the square root:

\sqrt{4 - x} = y - 3

, with

y \geq 3

. Under this condition, squaring gives

4 - x = {(y - 3)}^{2}

Let's solve in

ℝ

4 - x = {(y - 3)}^{2}

ⓘ We gather the terms in

x

on the right and the other terms on the left

4 - {(y - 3)}^{2} = x

ⓘ Raising a sum to an integer power is multiplying it by itself:

{(y - 3)}^{2}

(y - 3) \cdot (y - 3)

4 - (y - 3) \cdot (y - 3) = x

ⓘ Expand by multiplying each term inside the parentheses:

- (y - 3) \cdot (y - 3)

- (y \cdot y

+ y \cdot (-3)

- 3 y

- 3 \cdot (-3))

4 - (y \cdot y

+ y \cdot (-3)

- 3 y

- 3 \cdot (-3)) = x

ⓘ A minus sign before a parenthesis changes the sign of each of its terms:

- (y \cdot y

+ y \cdot (-3)

- 3 y

- 3 \cdot (-3))

- y \cdot y

- y \cdot (-3)

+ 3 y

+ 3 \cdot (-3)

4

- y \cdot y

- y \cdot (-3)

+ 3 y

+ 3 \cdot (-3) = x

ⓘ Multiplying powers with the same base adds the exponents:

y \cdot y

y^{2}

■Two minus signs cancel out:

- y \cdot (-3)

3 y

■Multiplications are done before additions and subtractions. A single minus sign does not cancel out, the product of a negative number by a positive number is negative:

3 \cdot (-3)

-9

4

- y^{2}

+ 3 y

+ 3 y

- 9 = x

ⓘ Start from

4

and move back

9

on the number line, going below zero:

4 - 9

-5

■Like terms are grouped by adding their coefficients:

3 y + 3 y

6 y

- 5 - y^{2} + 6 y = x

Solution obtained:

x = - y^{2} + 6 y - 5

Renaming the output as

x

, the inverse is

- x^{2} + 6 x - 5

, with domain

x \geq 3

▶Inverse function:

(- x^{2} + 6 x - 5 for x \geq 3)