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

The inverse of f(x)=(x^2-x)/(x+1) is (1+x+sqrt(x^2+6x+1))/2 ,(1+x-sqrt(x^2+6x+1))/2

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

inverse

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

\frac{1 + x + x ^{2} + 6 x + 1}{2}, \frac{1 + x - x ^{2} + 6 x + 1}{2}

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{1 + x + x ^{2} + 6 x + 1}{2}, \frac{1 + x - x ^{2} + 6 x + 1}{2}

f (x) = (0.5)

f (x) = - 2 x - 13

lo g_{3} (x + 3) - 2

f (x) = 3 x - 2 + 5

f (x) = (0.2)

What is the inverse of f(x)=(x^2-x)/(x+1) ?
The inverse of f(x)=(x^2-x)/(x+1) is (1+x+sqrt(x^2+6x+1))/2 ,(1+x-sqrt(x^2+6x+1))/2