What is the inverse of f(x)= 1/(sqrt(9-x)-3) ?

The inverse of f(x)= 1/(sqrt(9-x)-3) is -(1+6x)/(x^2)

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inverse of f(x)= 1/(sqrt(9-x)-3)

inverse

f (x) = \frac{1}{9 - x - 3}

- \frac{1 + 6 x}{x ^{2}}

Solution steps

y = \frac{1}{9 - x - 3}

Replace

x

with

y

x = \frac{1}{9 - y - 3}

Solve for

y, x = \frac{1}{9 - y - 3}

- \frac{1 + 6 x}{x ^{2}}

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

x^{2} - 3 x - 4

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

2 x + 11

4 x - x

What is the inverse of f(x)= 1/(sqrt(9-x)-3) ?
The inverse of f(x)= 1/(sqrt(9-x)-3) is -(1+6x)/(x^2)