Find the minimum and maximum values of the function $y=\left(t-3 t^{2}\right)^{1 / 3}$ on the interval $[-1,5]$ by comparing values at the critical points and endpoints.
(Use decimal notation, Give your answers to three decimal places.)
minimum:
maximum:

Step 1 ：Find the derivative of the function \(y=(t-3t^2)^{1/3}\), which is \(y' = \frac{1}{3}(t-3t^2)^{-2/3}(1-6t)\)

Step 2 ：Set the derivative equal to zero: \(\frac{1}{3}(t-3t^2)^{-2/3}(1-6t) = 0\)

Step 3 ：Solve for \(t\) to find the critical points: \(t = \frac{1}{6}\)

Step 4 ：Check where the derivative is undefined by setting the denominator of the derivative equal to zero: \(t-3t^2 = 0\)

Step 5 ：Solve for \(t\) to find the critical points: \(t = 0\) and \(t = \frac{1}{3}\)

Step 6 ：Evaluate the function at the critical points and at the endpoints of the interval \([-1,5]\): \(y(-1) = ((-1)-3(-1)^2)^{1/3} = -1.442\), \(y(0) = (0-3(0)^2)^{1/3} = 0\), \(y(\frac{1}{6}) = ((\frac{1}{6})-3(\frac{1}{6})^2)^{1/3} = 0.211\), \(y(\frac{1}{3}) = ((\frac{1}{3})-3(\frac{1}{3})^2)^{1/3} = 0\), and \(y(5) = (5-3(5)^2)^{1/3} = -7.368\)

Step 7 ：Compare these values to find the minimum and maximum values of the function on the interval \([-1,5]\)

Step 8 ：The minimum value of the function on the interval \([-1,5]\) is \(\boxed{-7.368}\)

Step 9 ：The maximum value of the function on the interval \([-1,5]\) is \(\boxed{0.211}\)