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

The inverse of f(x)= 1/3 (x+1)^2-2 is sqrt(3x+6)-1,-sqrt(3x+6)-1

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

inverse

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

3 x + 6 - 1, - 3 x + 6 - 1

Solution steps

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

Replace

x

with

y

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

Solve for

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

3 x + 6 - 1, - 3 x + 6 - 1

f (x) = \frac{x ^{2}}{16}

f (x) = x^{3} - 2 x^{2} - 15 x + 10

f (x) = x^{3} + 2 x^{2} + x - 7

x (n) = cos (3 n)

f (x) = \frac{x ^{2} + 12}{x - 2}

What is the inverse of f(x)= 1/3 (x+1)^2-2 ?
The inverse of f(x)= 1/3 (x+1)^2-2 is sqrt(3x+6)-1,-sqrt(3x+6)-1