What is the inverse of 1/(sqrt(x^2-4x-5)) ?

The inverse of 1/(sqrt(x^2-4x-5)) is (2x+sqrt(9x^2+1))/x ,(2x-sqrt(9x^2+1))/x

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inverse of 1/(sqrt(x^2-4x-5))

inverse

\frac{1}{x ^{2} - 4 x - 5}

\frac{2 x + 9 x ^{2} + 1}{x}, \frac{2 x - 9 x ^{2} + 1}{x}

Solution steps

y = \frac{1}{x ^{2} - 4 x - 5}

Replace

x

with

y

x = \frac{1}{y ^{2} - 4 y - 5}

Solve for

y, x = \frac{1}{y ^{2} - 4 y - 5}

\frac{2 x + 9 x ^{2} + 1}{x}, \frac{2 x - 9 x ^{2} + 1}{x}

f (x) = 16 x - 19

\frac{3 x}{7 - x}

f (x) = - 2 x^{3} - 7

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

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

What is the inverse of 1/(sqrt(x^2-4x-5)) ?
The inverse of 1/(sqrt(x^2-4x-5)) is (2x+sqrt(9x^2+1))/x ,(2x-sqrt(9x^2+1))/x