inverse of f(x)=4sqrt(2x+4)-3 — step-by-step solution

The function is strictly monotone on

x \geq -2

: it has an inverse on its range

y \geq -3

Isolate the square root:

\sqrt{2 x + 4} = \frac{y + 3}{4}

, with

y \geq -3

. Under this condition, squaring gives

2 x + 4 = {(\frac{y + 3}{4})}^{2}

Let's solve in

ℝ

2 x + 4 = {(\frac{y + 3}{4})}^{2}

ⓘ We subtract

4

from both sides

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

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

{(\frac{y + 3}{4})}^{2}

\frac{{(y + 3)}^{2}}{4^{2}}

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

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

4^{2}

4 \cdot 4

16

2 x = \frac{{(y + 3)}^{2}}{16} - 4

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

{(y + 3)}^{2}

(y + 3) \cdot (y + 3)

2 x = \frac{(y + 3) \cdot (y + 3)}{16} - 4

ⓘ Expand by multiplying each term inside the parentheses:

(y + 3) \cdot (y + 3)

y \cdot y

+ y \cdot 3

+ 3 y

+ 3 \cdot 3

2 x = \frac{y \cdot y + y \cdot 3 + 3 y + 3 \cdot 3}{16} - 4

ⓘ Multiplying powers with the same base adds the exponents:

y \cdot y

y^{2}

■Like terms are grouped by adding their coefficients:

y \cdot 3 + 3 y

6 y

■Multiplications are done before additions and subtractions:

3 \cdot 3

9

2 x = \frac{y^{2} + 6 y + 9}{16} - 4

ⅰ

\frac{y^{2} + 6 y + 9}{16} - 4

\frac{(y^{2} + 6 y + 9) + 16 \cdot (-4)}{16}

2 x = \frac{(y^{2} + 6 y + 9) + 16 \cdot (-4)}{16}

ⓘ 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:

16 \cdot (-4)

- 16 \cdot 4

- (10 + 6) \cdot 4

- (10 \cdot 4 + 6 \cdot 4)

- (40 + 24)

-64

2 x = \frac{y^{2} + 6 y + 9 - 64}{16}

ⓘ The numerator and the denominator are computed before the division:

9 - 64

-55

2 x = \frac{y^{2} + 6 y - 55}{16}

ⓘ Both sides are divided by

2

x = \frac{\frac{y^{2} + 6 y - 55}{16}}{2}

ⓘ Dividing a fraction by

2

is multiplying its denominator by

2

\frac{\frac{y^{2} + 6 y - 55}{16}}{2}

\frac{y^{2} + 6 y - 55}{16 \cdot 2}

x = \frac{y^{2} + 6 y - 55}{16 \cdot 2}

ⅰ

16 \cdot 2

(10 + 6) \cdot 2

10 \cdot 2 + 6 \cdot 2

20 + 12

32

Solution found:

x = \frac{y^{2} + 6 y - 55}{32}

Renaming the output as

x

, the inverse is

\frac{x^{2} + 6 x - 55}{32}

, with domain

x \geq -3

▶Inverse function:

(\frac{x^{2} + 6 x - 55}{32} for x \geq -3)