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

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

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

inverse

f (x) = (1 + x)^{2} + 2 x^{2}

\frac{- 1 + 3 x - 2}{3}, - \frac{3 x - 2 + 1}{3}

Solution steps

y = (1 + x)^{2} + 2 x^{2}

Replace

x

with

y

x = (1 + y)^{2} + 2 y^{2}

Solve for

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

\frac{- 1 + 3 x - 2}{3}, - \frac{3 x - 2 + 1}{3}

\frac{9}{2}

\frac{1}{x + 5} - 1

4 x^{2} + 5, x \geq 0

\frac{x}{5 x - 4}

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

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