92 CHAPTER 2 Functions and Their Graphs In calculus, there is a theorem with conditions that guarantee a function will have an absolute maximum and an absolute minimum. Solution (a) The function f whose graph is given in Figure 32(a) has the closed interval [ ] 0, 5 as its domain. The largest value of f is ( ) = f 3 6, the absolute maximum. The smallest value of f is ( ) = f 0 1, the absolute minimum. The function has a local maximum value of 6 at = x 3 and a local minimum value of 4 at = x 4. (b) The function f whose graph is given in Figure 32(b) has domain { } ≤ ≤ ≠ x x x 1 5, 3 . Note that we exclude 3 from the domain because of the “hole” at ( ) 3, 1 . The largest value of f on its domain is ( ) = f 5 3, the absolute maximum. There is no absolute minimum. Do you see why? As you trace the graph, getting closer to the point ( ) 3, 1 , there is no single smallest value. [As soon as you claim a smallest value, we can trace closer to ( ) 3, 1 and get a smaller value!] The function has no local maxima or minima. Remember that local extrema cannot occur at an endpoint. (c) The function f whose graph is given in Figure 32(c) has the interval [ ] 0, 5 as its domain. The absolute maximum of f is ( ) = f 5 4. The absolute minimum is f 2 1 ( ) = . The function has a local minimum value of 1 at x 2 = but no local maximum. (d) The function f given in Figure 32(d) has the interval [ )∞ 0, as its domain. The function has no absolute maximum; the absolute minimum is ( ) = f 0 0. The function has no local maxima or local minima. (e) The function f in Figure 32(e) has domain { } < < ≠ x x x 1 5, 2 . The function has no absolute maximum and no absolute minimum. Do you see why? The function has a local maximum value of 3 at = x 4, but no local minimum value. Finding the Absolute Maximum and the Absolute Minimum from the Graph of a Function For the graph of each function ( ) = y f x in Figure 32, find the absolute maximum and the absolute minimum, if they exist.Also, find any local maxima or local minima. EXAMPLE 5 Figure 32 (0, 1) (4, 4) (3, 6) (5, 5) 1 3 5 (a) x y 6 4 2 (3, 1) (1, 2) (5, 3) 1 3 5 (b) x y 6 4 2 (2, 1) (0, 3) (5, 4) 1 3 5 (c) x y 6 4 2 (0, 0) 1 3 5 (d) x y 6 4 2 (2, 2) (1, 4) (4, 3) 1 3 5 (e) x y 6 4 2 THEOREM Extreme Value Theorem If a function f is continuous* on a closed interval a b , , [ ] then f has an absolute maximum and an absolute minimum on a b , . [ ] The absolute maximum (minimum) can be found by selecting the largest (smallest) value of f from the following list: • The values of f at any local maxima or local minima of f in ( ) a b , . • The value of f at each endpoint of [ ] a b , —that is, ( ) f a and ( ) f b . *Although a precise definition requires calculus, we’ll agree for now that a function is continuous if its graph has no gaps or holes and can be traced without lifting the pencil from the paper. CAUTION A function may have an absolute maximum or an absolute minimum at an endpoint but not a local maximum or a local minimum. Why? Local maxima and local minima are found in an open interval I, and an open interval cannot be created around an endpoint. j
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