Find the absolute maximum and minimum of f on the interval [0,3], f(x)= x^3-4x^2+7x

To do this, find the first derivative of the given f(x) and set it = to 0.  Solve for the critical values:

f '(x) = 3x^2 - 4x + 7 = 0.  This has no real roots.  So there is neither relative max nor min.  However, we can find the max of f(x) on the interval [0,3] by calculating the y-values corresponding to {0, 1, 2, 3}:  {7, 6, 11, 39).
It would be wise to calculate more y-values.
However, with the y-values calculated for 0 and 3, we can venture a guess that the absolute minimum of this fn. on the given int is 7 and the abs. max is 39.

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