What is the inverse of (3x)/(sqrt(x^2-1)) ?

The inverse of (3x)/(sqrt(x^2-1)) is (xsqrt(x^2-9))/(x^2-9),-(xsqrt(x^2-9))/(x^2-9)

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inverse of (3x)/(sqrt(x^2-1))

inverse

\frac{3 x}{x ^{2} - 1}

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

Solution steps

y = \frac{3 x}{x ^{2} - 1}

Replace

x

with

y

x = \frac{3 y}{y ^{2} - 1}

Solve for

y, x = \frac{3 y}{y ^{2} - 1}

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

f (x) = 7 x + 5 y = 2

f (x) = \frac{3}{x - 8}

f (x) = 7 x + 3

f (x) = 7 x + 6

8 x + 5

What is the inverse of (3x)/(sqrt(x^2-1)) ?
The inverse of (3x)/(sqrt(x^2-1)) is (xsqrt(x^2-9))/(x^2-9),-(xsqrt(x^2-9))/(x^2-9)