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

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

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

inverse

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

- \frac{1}{x}, - - \frac{1}{x}

Solution steps

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

Replace

x

with

y

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

Solve for

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

- \frac{1}{x}, - - \frac{1}{x}

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

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

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

x^{2} + 2 x - 15

R (x) = \frac{x ^{2} - x - 12}{x + 6}

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