Absolute Extrema vs Relative Extrema Relative maximum A

Absolute Extrema vs Relative Extrema Relative maximum A peak, but not necessarily the highest number on the specified interval. Highest number on the specified interval, but not necessarily a peak. Relative minimum ALWAYS LEARNING Slide 1

In English: On a continuous, closed interval there must be a highest point and a lowest point. ALWAYS LEARNING Slide 2

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Example: Solution: Consider the function (a) Without graphing, show that f has an absolute minimum on the interval (0, 2). The derivative is which is defined everywhere, so the critical numbers are the solutions of The only critical number in the interval (0, 2) is Use the second-derivative test to determine whethere is a local extremum at Hence, f has a local minimum at Therefore, by the criticalpoint theorem, the absolute minimum of f on the interval (0, 2) occurs at ALWAYS LEARNING Slide 4

Example: Consider the function (b) Confirm the result of part (a) by graphing Solution: The graph of f in Figure 12. 38 shows that on the graph in the interval (0, 2). ALWAYS LEARNING is the lowest point Slide 5

For q Find the Absolute Maximum and Absolute Minimum on the closed interval [0, 5] so x = 4 is a critical point. q Evaluate f (0) , f (4), f (5) (the endpoints of the interval and any critical values between them) q qf (0) = – 02 + 8(0) – 12 = – 12 absolute minimum q f (4) = – 42 + 8(4) – 12 = 4 absolute maximum q f (5) = – 52 + 8(5) – 12 = 3 ALWAYS LEARNING Slide 6

[0, 5] The graph confirms the result ALWAYS LEARNING Slide 7

For q Find the Absolute Maximum and Absolute Minimum on the closed interval [0, 4] so is a critical point but is not on the specified interval. q Evaluate f (0) & f (4) (the endpoints of the interval only; there are no critical values between them) q qf (0) = 02 + (0) – 12 = – 12 absolute minimum q f (4) = 42 + (4) – 12 = 8 absolute maximum ALWAYS LEARNING Slide 8

[0, 4] Notice the graph gets lower at -½ , but not on that interval. The graph confirms the result ALWAYS LEARNING Slide 9