Thread: Prove that function is inverse of itself?

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

Originally Posted by fardeen_gen

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

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

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

Originally Posted by fardeen_gen

Show that the function given by $\displaystyle f(x) = \frac{k}{x},\ f:\mathbb{R} - \{0\}\rightarrow \mathbb{R} - \{0\}$, where $\displaystyle k\in\mathbb{R}, k\neq 0$ 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))=$\displaystyle f\left(\frac{k}{x}\right)=\frac{k}{\frac{k}{x}}=\f rac{k}{\frac{k}{x}}*\frac{x}{x}=\frac{kx}{k}=x$

Originally Posted by Sampras

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

No
We want(ed) to prove it...