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

The inverse of f(x)= 1/(sqrt(x^2+x-2)) is (-x+sqrt(9x^2+4))/(2x),-(x+sqrt(9x^2+4))/(2x)

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

inverse

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

\frac{- x + 9 x ^{2} + 4}{2 x}, - \frac{x + 9 x ^{2} + 4}{2 x}

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{- x + 9 x ^{2} + 4}{2 x}, - \frac{x + 9 x ^{2} + 4}{2 x}

f (x) = 1 + 6 + 9 x

f (x) = 7 - 2 x - 5

11

What is the inverse of f(x)= 1/(sqrt(x^2+x-2)) ?
The inverse of f(x)= 1/(sqrt(x^2+x-2)) is (-x+sqrt(9x^2+4))/(2x),-(x+sqrt(9x^2+4))/(2x)