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}

y = 5^{2 x}

\frac{1}{2 x}

f (x) = - 3 + x^{3}

f (x) = 6 x^{\frac{1}{3}} - 2

f (x) = \frac{4 x}{9 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