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.