What is the inverse of f(x)=(1/8)x^2+2x+7 ?

The inverse of f(x)=(1/8)x^2+2x+7 is (-16+sqrt(32x+32))/2 ,(-16-sqrt(32x+32))/2

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inverse of f(x)=(1/8)x^2+2x+7

inverse

f (x) = (\frac{1}{8}) x^{2} + 2 x + 7

\frac{- 16 + 32 x + 32}{2}, \frac{- 16 - 32 x + 32}{2}

Solution steps

y = \frac{1}{8} x^{2} + 2 x + 7

Replace

x

with

y

x = \frac{1}{8} y^{2} + 2 y + 7

Solve for

y, x = \frac{1}{8} y^{2} + 2 y + 7

\frac{- 16 + 32 x + 32}{2}, \frac{- 16 - 32 x + 32}{2}

f (x) = - 4 x^{9} - 4

f (x) = 5 x^{2} - 42

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

\frac{3}{3 x - 18}

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

What is the inverse of f(x)=(1/8)x^2+2x+7 ?
The inverse of f(x)=(1/8)x^2+2x+7 is (-16+sqrt(32x+32))/2 ,(-16-sqrt(32x+32))/2