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{2}{3} e (3 x - 1)

f (x) = 2 (x + 7)

f (x) = \frac{2 ^{x} + 8}{3 - 2 ^{x}}

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

f (x) = \frac{4}{x - 3} + 2

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)