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

The inverse of f(x)=x^2+x-1 is (-1+sqrt(4x+5))/2 ,(-1-sqrt(4x+5))/2

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

inverse

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

\frac{- 1 + 4 x + 5}{2}, \frac{- 1 - 4 x + 5}{2}

Solution steps

y = x^{2} + x - 1

Replace

x

with

y

x = y^{2} + y - 1

Solve for

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

\frac{- 1 + 4 x + 5}{2}, \frac{- 1 - 4 x + 5}{2}

f (x) = x^{2} - 2 x - 2

y = - tan (x - \frac{π}{2})

\frac{x ^{2} - 5 x + 6}{x ^{2} - 4 x + 3}

2 x - x^{2} + 8

f (x) = x^{2} + 6 x - 7

What is the inverse of f(x)=x^2+x-1 ?
The inverse of f(x)=x^2+x-1 is (-1+sqrt(4x+5))/2 ,(-1-sqrt(4x+5))/2