What is the inverse of f(x)=(25)/(x^2),x>0 ?

The inverse of f(x)=(25)/(x^2),x>0 is (5sqrt(x))/x ,-(5sqrt(x))/x

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inverse of f(x)=(25)/(x^2),x>0

inverse

f (x) = \frac{25}{x ^{2}}, x > 0

\frac{5 x}{x}, - \frac{5 x}{x}

Solution steps

y = \frac{25}{x ^{2}}

Replace

x

with

y

x = \frac{25}{y ^{2}}

Solve for

y, x = \frac{25}{y ^{2}}

\frac{5 x}{x}, - \frac{5 x}{x}

10 cos (\frac{2}{5} x)

f (x) = (x - 11)^{2}, [11, \infty]

- 6 x^{3} + 8

8 + 2 x - 2

What is the inverse of f(x)=(25)/(x^2),x>0 ?
The inverse of f(x)=(25)/(x^2),x>0 is (5sqrt(x))/x ,-(5sqrt(x))/x