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) = 82 - 7 ln (x + 2)

x + 2 + 5

x + 2 + 2

+ x^{2} + 1

y = 3 x^{2} - 5 x + 9

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