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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

\frac{x - 4}{x ^{2} + 8 x + 16}

f (x) = \frac{7}{4 x + 5}

q (x) = - 2 (- 3 x + 2)

f (x) = 2 e^{3 x^{2}}

y = lo g_{6} (3 x - 1) = 2

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