Range of re-parameterized function

Question

I am little confused here.

Let us define a function $f:[0,1]\rightarrow \mathbb{R}$, where

$\begin{equation*} \mathop{f}(t)= \left\{ \begin{array}{ll} 2\sqrt t & \mbox{if}\; 0\leq t < \frac{1}{2},\\ 5\sqrt t & \mbox{if}\; \frac{1}{2}\leq t \leq 1. \end{array} \right. \end{equation*} $

Then the re-parameterized function with $t\mapsto t^2$, will be

$\begin{equation*} \mathop{f}(t^2)= \left\{ \begin{array}{ll} 2t & \mbox{if}\; 0\leq t < \sqrt \frac{1}{2},\\ 5t & \mbox{if}\;\sqrt \frac{1}{2}\leq t \leq 1. \end{array} \right. \end{equation*} $

Could anyone please tell me is it right or not. I have doubt about the range of the re-parameterized function.

Thanks in advance.

Answer 1

This looks good to me.

$f(t)$ maps $[0,1/2)$ to $[0,\sqrt{2})$ and $[1/2,1]$ to $[5/\sqrt{2},5]$.

$f(t^2)$ maps $[0,\sqrt{1/2})$ to $[0,\sqrt{2})$ and $[\sqrt{1/2},1]$ to $[5/\sqrt{2},5].$

So, the range is the same.