5 Calculus Questions

1.

Find the absolute maximum and absolute minimum values of f on the given interval. f(x) = 6x3 - 18x2 - 54x + 7, [-2, 4] absolute minimum absolute maximumFind the absolute maximum and absolute minimum values of f on the given interval. f ( x) = x+ 25 x , [0.2, 20] absolute minimum value absolute maximum valueFind the absolute maximum and absolute minimum values of fon the given interval. f(t) = W 64 t2, [1: 3] E S absolute minimum value absolute maximum value EXAMPLE 4 The graph of the function f(x) = 3x4 - 28x3 + 60x2 -1 5 xs 6 (-1, 91) is shown in the figure. You can see that f(2) = is a local (2, 64) maximum whereas the absolute maximum is f( X -2 2 4 (This absolute maximum is not a local maximum because it occurs at an endpoint.) Also, f(0) = is a local minimum and f(5) = is both a local and an absolute minimum. Note that f has neither a local (5, -125) nor an absolute maximum at x = 6. Video Example ()Suppose you are given a formula for a function f. (a) How do you determine where f is increasing or decreasing? If f'(x) ? C 0 on an interval, then f is increasing on that interval. If f' ( x) ? 20 on an interval, then f is decreasing on that interval. i.. (b) How do you determine where the graph of f is concave upward or concave downward? If f"(x) ? C 0 for all x in I, then the graph of f is concave upward on I. If f"(x) ? C 0 for all x in I, then the graph of f is concave downward on I. (c) How do you locate inflection points? At any value of x where the concavity changes, we have an inflection point at (x, f(x)). At any value of x where the function changes from increasing to decreasing, we have an inflection point at (x, f( x) ). "At any value of x where the function changes from decreasing to increasing, we have an inflection point at (x, f(x) ). At any value of x where f'(x) = 0, we have an inflection point at (x, f(x)). At any value of x where the concavity does not change, we have an inflection point at (x, f(x))