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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

f (x) = \frac{x ^{2} - 7}{x - 4}

f (x) = \frac{1}{4 - x}

f (x) = \frac{∣ x ∣ - 2}{2 x + 2}

f (x) = e^{x} - x

f (x) = lo g_{2} (x - 3)

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