What is the inverse of f(x)= 1/(x^2),x<0 ?

The inverse of f(x)= 1/(x^2),x<0 is sqrt(1/x),-sqrt(1/x)

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inverse of f(x)= 1/(x^2),x<0

inverse

f (x) = \frac{1}{x ^{2}}, x < 0

\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}

f (x) = \frac{6}{( x + 5 )}

y = \frac{5 x - 8}{9 x + 6}

53. 7^{x - 0.7} + 2

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

1 - \frac{1}{x ^{3}}

What is the inverse of f(x)= 1/(x^2),x<0 ?
The inverse of f(x)= 1/(x^2),x<0 is sqrt(1/x),-sqrt(1/x)