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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

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

f (x) = 6 (x^{- \frac{1}{2}})

\frac{1}{( s ^{4} + 5 s ^{2} + 4 )}

x^{2} - 2 x + 3

f (x) = \frac{7 x + 1}{5 x + 2}

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