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

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

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

inverse

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

\frac{- 1 + 1 + 8 x}{2}, \frac{- 1 - 1 + 8 x}{2}

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{- 1 + 1 + 8 x}{2}, \frac{- 1 - 1 + 8 x}{2}

x^{2} 5 + x

5 x - 3 y = 12

f (x) = - 4 x + 9

- x + 5

g (x) = \frac{x}{5 x ^{2} + 4 x - 1}

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