What is the inverse of f(x)= 1/(sqrt(4x-3-x^2)) ?

The inverse of f(x)= 1/(sqrt(4x-3-x^2)) is -(-2x+sqrt(x^2-1))/x ,(2x+sqrt(x^2-1))/x

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inverse of f(x)= 1/(sqrt(4x-3-x^2))

inverse

f (x) = \frac{1}{4 x - 3 - x ^{2}}

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

Solution steps

y = \frac{1}{4 x - 3 - x ^{2}}

Replace

x

with

y

x = \frac{1}{4 y - 3 - y ^{2}}

Solve for

y, x = \frac{1}{4 y - 3 - y ^{2}}

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

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

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

\frac{x + 4}{x + 10}

f (x) = - e^{- x} + 9

f (x) = - e^{- x} + 7

What is the inverse of f(x)= 1/(sqrt(4x-3-x^2)) ?
The inverse of f(x)= 1/(sqrt(4x-3-x^2)) is -(-2x+sqrt(x^2-1))/x ,(2x+sqrt(x^2-1))/x