What is the inverse of sqrt(-3x^2-4x+4) ?

The inverse of sqrt(-3x^2-4x+4) is -(2+sqrt(-3x^2+16))/3 ,-(2-sqrt(-3x^2+16))/3

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inverse of sqrt(-3x^2-4x+4)

inverse

- 3 x^{2} - 4 x + 4

- \frac{2 + - 3 x ^{2} + 16}{3}, - \frac{2 - - 3 x ^{2} + 16}{3}

Solution steps

y = - 3 x^{2} - 4 x + 4

Replace

x

with

y

x = - 3 y^{2} - 4 y + 4

Solve for

y, x = - 3 y^{2} - 4 y + 4

- \frac{2 + - 3 x ^{2} + 16}{3}, - \frac{2 - - 3 x ^{2} + 16}{3}

f (x) = 1 + (x + 2)^{2}

f (x) = \frac{- 2 x - 1}{x - 2}

f (x) = e^{9 s i n (x)}

f (x) = \frac{- 5 x}{4} - 20

f (x) = x_{8}

What is the inverse of sqrt(-3x^2-4x+4) ?
The inverse of sqrt(-3x^2-4x+4) is -(2+sqrt(-3x^2+16))/3 ,-(2-sqrt(-3x^2+16))/3