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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

h (x) = \frac{4}{x - 5}

y = 4 x + 2, (- 8, 5)

sin^{2} (16 x)

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

2 x + 5 y = 8

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