What is the inverse of 3/(sqrt(1+x^2)) ?

The inverse of 3/(sqrt(1+x^2)) is (sqrt(9-x^2))/x ,-(sqrt(9-x^2))/x

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inverse of 3/(sqrt(1+x^2))

inverse

\frac{3}{1 + x ^{2}}

\frac{9 - x ^{2}}{x}, - \frac{9 - x ^{2}}{x}

Solution steps

y = \frac{3}{1 + x ^{2}}

Replace

x

with

y

x = \frac{3}{1 + y ^{2}}

Solve for

y, x = \frac{3}{1 + y ^{2}}

\frac{9 - x ^{2}}{x}, - \frac{9 - x ^{2}}{x}

f (x) = \frac{7 x - 6}{x + 3}

f (x) = \frac{8}{7 + x}

7 e^{x} - 3

f (x) = \frac{3 x ^{2} - 5}{6}

f (x) = - 2^{2} + 8 x - 11

What is the inverse of 3/(sqrt(1+x^2)) ?
The inverse of 3/(sqrt(1+x^2)) is (sqrt(9-x^2))/x ,-(sqrt(9-x^2))/x