What is the inverse of f(x)= 1/3 ((x-4)/2)^2-1 ?

The inverse of f(x)= 1/3 ((x-4)/2)^2-1 is 2sqrt(3x+3)+4,-2sqrt(3x+3)+4

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inverse of f(x)= 1/3 ((x-4)/2)^2-1

inverse

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

2 3 x + 3 + 4, - 2 3 x + 3 + 4

Solution steps

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

Replace

x

with

y

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

Solve for

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

2 3 x + 3 + 4, - 2 3 x + 3 + 4

f (x) = \frac{2 x - 4}{4 x + 2}

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

f (x) = 20 e^{(\frac{1}{1690} I n \frac{1}{2})}

g (x) = 4 + \frac{7}{2} x

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

What is the inverse of f(x)= 1/3 ((x-4)/2)^2-1 ?
The inverse of f(x)= 1/3 ((x-4)/2)^2-1 is 2sqrt(3x+3)+4,-2sqrt(3x+3)+4