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

The inverse of (4x)/(sqrt(x^2-1)) is (xsqrt(x^2-16))/(x^2-16),-(xsqrt(x^2-16))/(x^2-16)

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

inverse

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

\frac{x x ^{2} - 16}{x ^{2} - 16}, - \frac{x x ^{2} - 16}{x ^{2} - 16}

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{x x ^{2} - 16}{x ^{2} - 16}, - \frac{x x ^{2} - 16}{x ^{2} - 16}

f (x) = 3 (x - 1)^{2} + 4

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

\frac{2 x - 5}{x - 7}

f (x) = - \frac{2}{5} x + 2

f (x) = 5 - 2 (x - 1)^{2}

What is the inverse of (4x)/(sqrt(x^2-1)) ?
The inverse of (4x)/(sqrt(x^2-1)) is (xsqrt(x^2-16))/(x^2-16),-(xsqrt(x^2-16))/(x^2-16)