What is the inverse of f(x)= 2/3 (x-1)^2-3 ?

The inverse of f(x)= 2/3 (x-1)^2-3 is sqrt((3x+9)/2)+1,-sqrt((3x+9)/2)+1

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inverse of f(x)= 2/3 (x-1)^2-3

inverse

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

\frac{3 x + 9}{2} + 1, - \frac{3 x + 9}{2} + 1

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{3 x + 9}{2} + 1, - \frac{3 x + 9}{2} + 1

f (x) = - 2 x - 3 - 1

f (x) = 140 (1.6)^{x}

f (x) = (x + 5)^{\frac{2}{7}}

sin (6 x)

f (x) = \frac{x + 6}{x ( x + 11 )}

What is the inverse of f(x)= 2/3 (x-1)^2-3 ?
The inverse of f(x)= 2/3 (x-1)^2-3 is sqrt((3x+9)/2)+1,-sqrt((3x+9)/2)+1