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

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

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inverse of f(x)= 1/(x^2-6x+9)

inverse

f (x) = \frac{1}{x ^{2} - 6 x + 9}

\frac{3 x + x}{x}, \frac{3 x - x}{x}

Solution steps

y = \frac{1}{x ^{2} - 6 x + 9}

Replace

x

with

y

x = \frac{1}{y ^{2} - 6 y + 9}

Solve for

y, x = \frac{1}{y ^{2} - 6 y + 9}

\frac{3 x + x}{x}, \frac{3 x - x}{x}

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

f (x) = 6 x - 11

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

f (x) = x^{5} + 2

f (x) = 2^{x + 1} - 3

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