Prove that function is inverse of itself?

**fardeen_gen** · June 25th 2009, 02:00 PM

Show that the function given by $f(x) = \frac{k}{x},\ f:\mathbb{R} - \{0\}\rightarrow \mathbb{R} - \{0\}$ , where $k\in\mathbb{R}, k\neq 0$ is inverse of itself.

**mr fantastic** · June 25th 2009, 07:29 PM

Originally Posted by fardeen_gen

Show that the function given by $f(x) = \frac{k}{x},\ f:\mathbb{R} - \{0\}\rightarrow \mathbb{R} - \{0\}$ , where $k\in\mathbb{R}, k\neq 0$ is inverse of itself.

How have you been taught to find the inverse of a function and where are you stuck in that procedure?

**Sampras** · June 25th 2009, 07:30 PM

$f(f(x)) = x$ e.g. it is the identity element. This is given that we know $f$ is its own inverse.

**artvandalay11** · June 25th 2009, 07:32 PM

Originally Posted by fardeen_gen

Show that the function given by $f(x) = \frac{k}{x},\ f:\mathbb{R} - \{0\}\rightarrow \mathbb{R} - \{0\}$ , where $k\in\mathbb{R}, k\neq 0$ is inverse of itself.

For two functions, g(x) and f(x), to be inverses of eachother, g(f(x))=x and f(g(x))=x

So if f(x) is it's own inverse, then f(f(x))=x and swapping the order, f(f(x))=x, which is the same identity.

So f(f(x))= $f\left(\frac{k}{x}\right)=\frac{k}{\frac{k}{x}}=\f rac{k}{\frac{k}{x}}*\frac{x}{x}=\frac{kx}{k}=x$

**Moo** · June 26th 2009, 12:02 AM

Originally Posted by Sampras

$f(f(x)) = x$ e.g. it is the identity element. This is given that we know $f$ is its own inverse.

No

We want(ed) to prove it...

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