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

The inverse of y=(x^2+2x+1)/(x+3) is (-2+x+sqrt(x^2+8x))/2 ,(-2+x-sqrt(x^2+8x))/2

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

inverse

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

\frac{- 2 + x + x ^{2} + 8 x}{2}, \frac{- 2 + x - x ^{2} + 8 x}{2}

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{- 2 + x + x ^{2} + 8 x}{2}, \frac{- 2 + x - x ^{2} + 8 x}{2}

f (x) = 0.2 x

(- 4, 0), (6, - 7)

f (x) = x^{2} - 7 x + 6

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

f (x) = π (x)^{2}

What is the inverse of y=(x^2+2x+1)/(x+3) ?
The inverse of y=(x^2+2x+1)/(x+3) is (-2+x+sqrt(x^2+8x))/2 ,(-2+x-sqrt(x^2+8x))/2