Well, |t-3| >= 0, becoming its minimum value (0) at t=3.
So, y has a maximum value of 3 at t=3, since it is subtracting 0 from 3. For anything other than 3, y is less than 3.
The other extrema would be minima, occurring at the ends of the interval.
Locate the absolute exterma of the function on the closed interval.
y=3-|t-3| , [-1,5]
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