inverse of sqrt(x-1)-4 — step-by-step solution

The function is strictly monotone on

x \geq 1

: it has an inverse on its range

y \geq -4

Isolate the square root:

\sqrt{x - 1} = y + 4

, with

y \geq -4

. Under this condition, squaring gives

x - 1 = {(y + 4)}^{2}

⇥Let's solve in

ℝ

x - 1 = {(y + 4)}^{2}

⇥ⓘ We add

1

to both sides

⇥

x = {(y + 4)}^{2} + 1

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

{(y + 4)}^{2}

(y + 4) \cdot (y + 4)

⇥

x = (y + 4) \cdot (y + 4) + 1

⇥ⓘ Expand by multiplying each term inside the parentheses:

(y + 4) \cdot (y + 4)

y \cdot y

+ y \cdot 4

+ 4 y

+ 4 \cdot 4

⇥

x = y \cdot y

+ y \cdot 4

+ 4 y

+ 4 \cdot 4

+ 1

⇥ⓘ Multiplying powers with the same base adds the exponents:

y \cdot y

y^{2}

■Like terms are grouped by adding their coefficients:

y \cdot 4 + 4 y

8 y

■Multiplications are done before additions and subtractions:

4 \cdot 4

16

⇥

x = y^{2}

+ 8 y

+ 16

+ 1

⇥ⓘ Start from

16

and move forward

1

on the number line:

16 + 1

17

⇥Solution obtained:

x = y^{2} + 8 y + 17

Renaming the output as

x

, the inverse is

x^{2} + 8 x + 17

, with domain

x \geq -4

▶Inverse function:

(x^{2} + 8 x + 17 for x \geq -4)