Consider a domain $\Omega = [0, 1]$ and some stepsize $d\in \mathbb{N}$. We create an equidistant partition $a_0 := 0$, $a_\ell := a_{\ell - 1} + \frac{1}{d}$ for $\ell \in \lbrace 1, ..., d \rbrace$.
We define piecewise cubic polynomials $\phi_i: \Omega \rightarrow \mathbb{R}$ where $i = 0,1, ..., 2d+1$ by the rule $$ \phi_i(a_\ell) = \begin{cases} 1,~ \frac{i}{2} = \ell \\ 0,~ \text{otherwise} \end{cases} \quad \phi_i'(a_\ell) = \begin{cases} 1,~ \frac{i-1}{2} = \ell \\ 0,~ \text{otherwise} \end{cases} $$ as $\ell = 0, ..., d$.
Then I was given the polynomials $$ \psi_1(x) = 2x^3 - 3x^2 +1, ~\psi_2(x) = x^3-2x^2+x, ~\psi_3(x) = -2x^3+3x^2, ~\psi_4(x) = x^3-x^2. $$ on $\Omega$. I also received a hint that there are four functions $\phi_i$ that are non-zero on every interval $[a_{\ell-1}, a_\ell]$ and they can (on their support) be obtained by "affine transformation" of some $\psi_i$. My question is: How does this affine transformation look like? And where do those $\psi_i$ come from?
I calculated $\phi_1\vert_{\left[\frac{1}{5}, \frac{2}{5}\right]}(x) = 250x^3-225x^2+60x-4 = \psi_1(5x-1)$ for the case $d = 5$... . Is this the desired relation and if yes, why?