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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

f (x) = x^{2} + y^{2} = 16

f (x) = 2 (- 1 + e^{t})

f (x) = \frac{5 x + 2}{9 x - 4}

\frac{6 x}{36 - x ^{2}}

lo g_{10} (x + 5)

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