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

The inverse of (1/2-x)^2-2 is -(2sqrt(x+2)-1)/2 ,-(-2sqrt(x+2)-1)/2

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inverse of (1/2-x)^2-2

inverse

(\frac{1}{2} - x)^{2} - 2

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

f (x) = \frac{( 7 x - 1 )}{9 x + 8}

f (x) = 4 - 7 y

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

x + 2 - 2 x + 3

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