For a problem I am working with a transformation of the function $g(t)=t^2$ using a piecewise function as follows:
$$f(t)=\begin{cases}(2t)^2,& \text{for }t\in[0,0.5],\\(-2t+2)^2,&\text{for }t\in(0.5,1].\end{cases}$$
I have been told that
$$f(t) = g(1-|2t-1|)$$
is a closed-form, however I cannot see how one can deduce this closed form algebraically given the piecewise function. How can this be done?
First note that $g(2t)$ equals the curve in the first part of the piecewise function. Next see that $g(2-2t)$ equals the curve in the second part of the piecewise function.
\begin{align*} f(t)&=\begin{cases}(2t)^2,& \text{for }t\in[0,0.5],\\(-2t+2)^2,&\text{for }t\in(0.5,1].\end{cases}\\ f(t)&=\begin{cases}g(2t),& \text{for }t\in[0,0.5],\\g(-2t+2),&\text{for }t\in(0.5,1].\end{cases}\\ f(t)&=\begin{cases}g(1+2t-1),& \text{for }t\in[0,0.5],\\g(1-(2t-1)),&\text{for }t\in(0.5,1].\end{cases}\\ f(t)&=\begin{cases}g(1-|2t-1|),& \text{for }t\in[0,0.5],\\g(1-|2t-1|),&\text{for }t\in(0.5,1].\end{cases}\\ \end{align*} Clearly, $f(t) = g(1-|2t-1|) \quad \forall t\in [0,1]$. The trick here is to notice in the second step that the coefficients of $2t$ have the same magnitude, but opposite sign (or equivalently, to notice in the image you pasted that there is a symmetry around $\frac{1}{2}$ for $2|t-\frac{1}{2}| = |2t-1|$)