What is the inverse of 9-9/(x^2),x>0 ?

The inverse of 9-9/(x^2),x>0 is sqrt(-9/(x-9)),-sqrt(-9/(x-9))

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inverse of 9-9/(x^2),x>0

inverse

9 - \frac{9}{x ^{2}}, x > 0

- \frac{9}{x - 9}, - - \frac{9}{x - 9}

Solution steps

y = 9 - \frac{9}{x ^{2}}

Replace

x

with

y

x = 9 - \frac{9}{y ^{2}}

Solve for

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

- \frac{9}{x - 9}, - - \frac{9}{x - 9}

f (x) = \frac{5 x}{6 x - 1}

f (x) = (x + 4)^{\frac{2}{7}}

f (x) = 2^{x - 4}

f (x) = ln (16 - t^{2})

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

What is the inverse of 9-9/(x^2),x>0 ?
The inverse of 9-9/(x^2),x>0 is sqrt(-9/(x-9)),-sqrt(-9/(x-9))