inverse of f(x)=2sqrt(x-5) — step-by-step solution

The function is strictly monotone on

x \geq 5

: it has an inverse on its range

y \geq 0

Isolate the square root:

\sqrt{x - 5} = \frac{y}{2}

, with

y \geq 0

. Under this condition, squaring gives

x - 5 = {(\frac{y}{2})}^{2}

⇥Let's solve in

ℝ

x - 5 = {(\frac{y}{2})}^{2}

⇥ⓘ We add

5

to both sides

⇥

x = {(\frac{y}{2})}^{2} + 5

⇥ⓘ The power of a quotient is the quotient of the powers, (a/b)^n = a^n/b^n:

{(\frac{y}{2})}^{2}

\frac{y^{2}}{2^{2}}

⇥

x = \frac{y^{2}}{2^{2}} + 5

⇥ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product:

2^{2}

2 \cdot 2

4

⇥Solution obtained:

x = \frac{y^{2}}{4} + 5

Renaming the output as

x

, the inverse is

\frac{x^{2}}{4} + 5

, with domain

x \geq 0

▶Inverse function:

(\frac{x^{2}}{4} + 5 for x \geq 0)