What is the inverse of f(x)=(7x)/(sqrt(x^2-1)) ?

The inverse of f(x)=(7x)/(sqrt(x^2-1)) is (xsqrt(x^2-49))/(x^2-49),-(xsqrt(x^2-49))/(x^2-49)

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inverse of f(x)=(7x)/(sqrt(x^2-1))

inverse

f (x) = \frac{7 x}{x ^{2} - 1}

\frac{x x ^{2} - 49}{x ^{2} - 49}, - \frac{x x ^{2} - 49}{x ^{2} - 49}

Solution steps

y = \frac{7 x}{x ^{2} - 1}

Replace

x

with

y

x = \frac{7 y}{y ^{2} - 1}

Solve for

y, x = \frac{7 y}{y ^{2} - 1}

\frac{x x ^{2} - 49}{x ^{2} - 49}, - \frac{x x ^{2} - 49}{x ^{2} - 49}

x + 3 - 1

f (x) = \frac{1 + 2 x}{7 + x}

f (x) = \frac{x ^{\frac{1}{3}}}{5} + 8

(x + 2)^{3} - 8

f (x) = \frac{x}{9} + 1

What is the inverse of f(x)=(7x)/(sqrt(x^2-1)) ?
The inverse of f(x)=(7x)/(sqrt(x^2-1)) is (xsqrt(x^2-49))/(x^2-49),-(xsqrt(x^2-49))/(x^2-49)