inverse of f(x)=-2sqrt(3x-1)+4 — step-by-step solution

The function is strictly monotone on

x \geq \frac{1}{3}

: it has an inverse on its range

y \leq 4

Isolate the square root:

\sqrt{3 x - 1} = \frac{y - 4}{-2}

, with

y \leq 4

. Under this condition, squaring gives

3 x - 1 = {(\frac{y - 4}{-2})}^{2}

Let's solve in

ℝ

3 x - 1 = {(\frac{y - 4}{-2})}^{2}

ⓘ The sign of a fraction goes in front of the bar, whether it comes from the numerator or the denominator:

\frac{y - 4}{-2}

- \frac{y - 4}{2}

■With an even integer exponent, the minus sign in the base disappears; with an odd integer exponent, it moves in front of the power:

{(- \frac{y - 4}{2})}^{2}

{(\frac{y - 4}{2})}^{2}

3 x - 1 = {(\frac{y - 4}{2})}^{2}

ⓘ We add

1

to both sides

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

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

{(\frac{y - 4}{2})}^{2}

\frac{{(y - 4)}^{2}}{2^{2}}

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

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

2^{2}

2 \cdot 2

4

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

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

{(y - 4)}^{2}

(y - 4) \cdot (y - 4)

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

3 x = \frac{y \cdot y + y \cdot (-4) - 4 y - 4 \cdot (-4)}{4} + 1

ⓘ Multiplying powers with the same base adds the exponents:

y \cdot y

y^{2}

■The minus sign is placed in front of the product:

y \cdot (-4)

- 4 y

■Multiplications are done before additions and subtractions. Two minus signs cancel out, the product of two negative numbers is positive:

-4 \cdot (-4)

16

3 x = \frac{y^{2} - 4 y - 4 y + 16}{4} + 1

ⓘ Like terms are grouped by adding their coefficients:

- 4 y - 4 y

-8 y

3 x = \frac{y^{2} - 8 y + 16}{4} + 1

ⅰ

\frac{y^{2} - 8 y + 16}{4} + 1

\frac{(y^{2} - 8 y + 16) + 4}{4}

3 x = \frac{(y^{2} - 8 y + 16) + 4}{4}

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

16 + 4

20

3 x = \frac{y^{2} - 8 y + 20}{4}

ⓘ A fraction whose numerator is a sum splits term by term, (a + b)/c = a/c + b/c:

\frac{y^{2} - 8 y + 20}{4}

- 2 y + \frac{y^{2} + 20}{4}

3 x = - 2 y + \frac{y^{2} + 20}{4}

ⓘ Both sides are divided by

3

x = \frac{- 2 y + \frac{y^{2} + 20}{4}}{3}

ⓘ To add fractions, put them over the same denominator, then add the numerators:

- 2 y + \frac{y^{2} + 20}{4}

\frac{- 2 y \cdot 4 + (y^{2} + 20)}{4}

x = \frac{\frac{- 2 y \cdot 4 + (y^{2} + 20)}{4}}{3}

ⓘ Multiplications are done before additions and subtractions:

2 \cdot 4

8

x = \frac{\frac{- 8 y + y^{2} + 20}{4}}{3}

ⓘ Dividing a fraction by

3

is multiplying its denominator by

3

\frac{\frac{- 8 y + y^{2} + 20}{4}}{3}

\frac{- 8 y + y^{2} + 20}{4 \cdot 3}

x = \frac{- 8 y + y^{2} + 20}{4 \cdot 3}

ⅰ

4 \cdot 3

12

x = \frac{- 8 y + y^{2} + 20}{12}

Solution obtained:

x = \frac{y^{2} - 8 y + 20}{12}

Renaming the output as

x

, the inverse is

\frac{x^{2} - 8 x + 20}{12}

, with domain

x \leq 4

▶Inverse function:

(\frac{x^{2} - 8 x + 20}{12} for x \leq 4)