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

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

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inverse of f(x)=x+x^2

inverse

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

\frac{- 1 + 1 + 4 x}{2}, \frac{- 1 - 1 + 4 x}{2}

Solution steps

y = x + x^{2}

Replace

x

with

y

x = y + y^{2}

Solve for

y, x = y + y^{2}

\frac{- 1 + 1 + 4 x}{2}, \frac{- 1 - 1 + 4 x}{2}

x^{2} + 2, x \leq 0

f (x) = 8 (2 x + 10)^{2} + 3, x \geq - 5

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

f (x) = \frac{( 4 x ^{2} - 3 )}{5}

f (x) = 6 + 3 x - 3

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