How do you find the inverse of $f (x) = \frac{1}{x^{2}}$ $f(x)=1/x^2$ ?

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How do you find the inverse of $f (x) = \frac{1}{x^{2}}$ $f(x)=1/x^2$ ?

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Answer 1 · 2015-11-30T22:42:35

$y = \pm \frac{1}{\sqrt{x}}$ $y=+-1/sqrtx$

Explanation:

Rewrite as $y = \frac{1}{x^{2}}$ $y=1/x^2$ .

Switch the $x$ $x$ and $y$ $y$ .

$x = \frac{1}{y^{2}}$ $x=1/y^2$

Solve for $y$ $y$ .

$x y^{2} = 1$ $xy^2=1$

$y^{2} = \frac{1}{x}$ $y^2=1/x$

$y = \pm \frac{1}{\sqrt{x}}$ $y=+-1/sqrtx$

Notice that the inverse is not actually a function, but two coexisting functions: $\frac{1}{\sqrt{x}}$ $1/sqrtx$ and $- \frac{1}{\sqrt{x}}$ $-1/sqrtx$ .

You can tell if a function's inverse will be a function if the function is one-to-one, or if it passes the "horizontal line test".

This is the graph of $f (x)$ $f(x)$ :
graph{1/x^2 [-10.25, 9.75, -2.04, 7.96]}

Notice how all $y$ $y$ -values are represented by two $x$ $x$ -values. Since the domain and range of the inverse function will flip, you can tell the inverse won't be a function.

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How do you find the inverse of $f (x) = \frac{1}{x^{2}}$ $f(x)=1/x^2$ ?

1 Answer

Explanation:

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How do you find the inverse of $f (x) = \frac{1}{x^{2}}$ $f(x)=1/x^2$ ?