What is the inverse of f(x)= 1/8 (1/2 x^2+2x+2) ?

The inverse of f(x)= 1/8 (1/2 x^2+2x+2) is -2+4sqrt(x),-2-4sqrt(x)

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inverse of f(x)= 1/8 (1/2 x^2+2x+2)

inverse

f (x) = \frac{1}{8} (\frac{1}{2} x^{2} + 2 x + 2)

- 2 + 4 x, - 2 - 4 x

Solution steps

y = \frac{1}{8} (\frac{1}{2} x^{2} + 2 x + 2)

Replace

x

with

y

x = \frac{1}{8} (\frac{1}{2} y^{2} + 2 y + 2)

Solve for

y, x = \frac{1}{8} (\frac{1}{2} y^{2} + 2 y + 2)

- 2 + 4 x, - 2 - 4 x

f (x) = 5 e^{y}

f (x) = \frac{x - 1}{x + 6}

f (x) = \frac{- x + 5}{2 x - 3}

+ 8 x - 7

f (x) = x (8 - x)

What is the inverse of f(x)= 1/8 (1/2 x^2+2x+2) ?
The inverse of f(x)= 1/8 (1/2 x^2+2x+2) is -2+4sqrt(x),-2-4sqrt(x)