For any function of the form \(y = f(t)\text{,}\) one way to determine if we can view the original input variable \(t\) as a function of the original output variable \(y\) is to attempt to solve the equation \(y = f(t)\) for \(t\) in terms of \(y\text{.}\)
Taking \(y = 3 - \frac{1}{5}(t-1)^3\text{,}\) we try to solve for \(t\) by first subtracting \(3\) from both sides to get
\begin{equation*}
y - 3 = -\frac{1}{5}(t-1)^3\text{.}
\end{equation*}
Next, multiplying both sides by \(-5\text{,}\) it follows that
\begin{equation*}
(t-1)^3 = -5(y-3)\text{.}
\end{equation*}
Because the cube root function has the property that \(\sqrt[3]{z^3} = z\) for every real number \(z\) (since the cube root function is the inverse function for the cubing function, and each function has both a domain and range of all real numbers), we can take the cube root of both sides of the preceding equation to get
\begin{equation*}
t - 1 = \sqrt[3]{-5(y-3)}\text{.}
\end{equation*}
Finally, adding \(1\) to both sides, we have determined that
\begin{equation*}
t = 1 + \sqrt[3]{-5(y-3)}\text{.}
\end{equation*}
Because we have been able to express \(t\) as a single function of \(y\) for every possible value of \(y\text{,}\) this shows that \(r\) indeed has an inverse and that \(t = r^{-1}(y) = 1 + \sqrt[3]{-5(y-3)}\text{.}\)
We attempt similar reasoning for the second function, \(y = 3 - \frac{1}{5}(t-1)^2\text{.}\) To solve for \(t\text{,}\) we first subtract \(3\) from both sides, so that
\begin{equation*}
y - 3 = -\frac{1}{5}(t-1)^2\text{.}
\end{equation*}
After multiplying both sides by \(-5\text{,}\) we have
\begin{equation*}
(t-1)^2 = -5(y-3)\text{.}
\end{equation*}
Next, it's necessary to take the square root of both sides in an effort to isolate \(t\text{.}\) Here, however, we encounter a crucial issue. Because the function \(g(x) = x^2\) takes any nonzero number and its opposite to the same output (e.g. \((-5)^2 = 25 = (5)^2\)), this means that we have to account for both possible inputs that result in the same output. Based on our last equation, this means that either
\begin{equation*}
t-1 = \sqrt{-5(y-3)} \text{ or } t-1 = -\sqrt{-5(y-3)}\text{.}
\end{equation*}
As such, we find not a single equation that expresses \(t\) as a function of \(y\text{,}\) but rather two:
\begin{equation*}
t = 1 + \sqrt{-5(y-3)} \text{ or } t = 1 -\sqrt{-5(y-3)}\text{.}
\end{equation*}
Since it appears that \(t\) can't be expressed as a single function of \(y\text{,}\) it seems to follow that \(y = s(t) = 3 - \frac{1}{5}(t-1)^2\) does not have an inverse function.