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Page 25 Applications of the DerivativesFinding Local/Relative Extrema Local (or Relative) Extrema A function f has a local maximum at x = x 0 if locally, f(x 0 ) is greater than all the surrounding values of f(x). We call this f(x 0 ) a local maximum of f. A function f has a local minimum at x = x 0 if locally, f(x 0 ) is less than all the surrounding values of f(x). We call this f(x 0 ) a local minimum of f. x1x1 x2x2 x3x3 f(x 3 ) is a __________ f(x 1 ) is a ________________________ f(x 2 ) is a __________ Global (or Absolute) Extrema A function f has a global maximum at x = x 0 if f(x 0 ) is greater than or equal to (i.e., the greatest) all values of f(x). We call this f(x 0 ) the global maximum of f. A function f has a global minimum at x = x 0 if f(x 0 ) is less than or equal to (i.e., the smallest) all values of f(x). We call this f(x 0 ) the global minimum of f. Q: What is the slope of the tangent line at a local/relative extremum such as the ones above? A: ______________________________________________ Therefore, to find all local/relative extrema (pl. of extremum) of a function f(x), we ask ourselves this question: For what x-value(s), ________? Examples: For each of the following functions, find all of its local/relative extrema and for each extremum, the x-values for which the extremum is obtained. 1. f(x) = 2x 2 – 4x f(x) = x 3 + 3x 2 – 9x f(x) = sin x

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Page 26 Bounded vs. Unbounded Two questions might come across your mind: Q1: Can we use the same method to find global/ absolute extrema? and Q2: How do we determine whether a extremum (weve found) is a maximum or a minimum? We will answer the first question only (and save the second one for later) and the answer is YES and NO. Its a YES because we still will use the idea of setting f (x) = 0. Its a NO because many functions do not have global/absolute extrema. For example, if we look at the graphs of the quadratic function f(x) = 2x 2 – 4x + 3, the cubic function f(x) = x 3 + 3x 2 – 9x + 6, and the sine function f(x) = sin x on the last page, we can see that the quadratic function has a _________ extremum point at (_, _), whereas the cubic function has only _________ extremum points at (_, _) and (_, _) and the sine function has _________ many _________ extremum points (despite it only has one global maximum, which is _, and one global minimum, which is __.) f(x) = sin x f(x) = x 3 + 3x 2 – 9x + 6 f(x) = 2x 2 – 4x + 3 We say a function f(x) is bounded above by A, if no function values of f is higher than A, i.e., f(x) A for all x in its domain. (See figure 4.) We say a function f(x) is bounded below by B, if no function values of f is lower than B, i.e, f(x) B for all x in its domain. (See figure 5.) Finally, we say a function f(x) is (simply) bounded if it is bounded above (by one number) and bounded below (by another number). If we say a function f(x) is bounded by M (where M > 0), it means f(x) is bounded above by M and bounded below by –M, i.e., ___________ for all x in its domain. (See figure 6.) Notice the three functions above are continuous everywhere (i.e., on all real numbers). Unless it is bounded (e.g., f(x) = sin x), then there will be a global max and global min. Otherwise, there will not be a global max and/or a global min (e.g., f(x) = 2x 2 – 4x + 3 has a global min but no global max, whereas f(x) = x 3 + 3x 2 – 9x + 6 has neither a global max nor a global min). Figure 4: f(x) is bounded above by 2; g(x) is bounded above by 3 2 y = f(x) 3 y = g(x) Figure 5: f(x) is bounded below by –2 –2 y = f(x) 3 –3 Figure 6: f(x) is bounded by 3

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Page 27 Application of DerivativesFinding the Global Max and Global Min If f(x) is continuous on a closed interval [a, b], the two global extrema will exist. They will occur at c (a < c < b) with either f (c) = 0 or f (c) does not exist, or at a, or at b. In other words, the global extrema will occur at some value(s) between a and b or at the endpoints a and b. The largest of f(a), f(b) and f(c)s will be the global maximum and the smallest of these values will the global minimum. Examples: Find the global maximum and global minimum of the following functions: 1. f(x) = 2x 2 – 4x + 3 on [–1, 2] 2. f(x) = x 3 + 3x 2 – 9x + 6 on [–3, 3]3. f(x) = x e x on [-2, 1] Now, lets say we know a function is continuous everywhere, but we dont know its graph (so we cant tell its bounded or not), how can we find its global extrema? We cant UNLESS we restrict f(x) on some closed interval, say [a, b]. If our everywhere-continuous function is restricted on a closed interval, not only we can find its global extrema, we can also determine whether an extremum is the maximum or the minimum (after all, an extremum can only mean a maximum or a minimum). This can be stated in the following theorem, called the Extreme Value Theorem: A word on maximum and minimum: 1. can never be considered as global maximum and – can never be considered as _global minimum_. 2.A global/local maximum of a function is the largest real y-value of the function globally/locally. Similarly, a global/local minimum of a function is the smallest real y-value of the function globally/locally. 3.A global/local maximum point is a point (x 0, y 0 ) with y 0 being the largest real y-value of the function globally/locally. Similarly for global/local minimum point. 4.The y-value of a hole can never be considered as maximum or minimum because the function can never reach there, despite of being very close. 5. The y-value of an isolated point can be considered as maximum/minimum –1 –2 2 –3 –4 –1–2–4–5–6 2346

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Page 28 The Extreme Value Theorem If f(x) is continuous on a closed interval [a, b], the two global extrema will exist. They will occur at c (a < c < b) with either f (c) = 0 or f (c) does not exist, or at a, or at b. In other words, the global extrema will occur at some value(s) between a and b or at the endpoints a and b. The largest of f(a), f(b) and f(c)s will be the global maximum and the smallest of these values will the global minimum. The continuous clause in the theorem is important because, if the function is not continuous, the global max/min might not be guaranteed. (See figure 7.) The closed interval clause is also important because, if the function is continuous on an opened interval (a, b) but not on the closed interval [a, b], the global max/min also might not be guaranteed. (See figure 8.) A good thing about the Extreme Value Theorem is that the function does not to be continuous everywhere (we apologize if we have previously given you the impression that it must be). As long as the function is continuous on the specified closed interval, the theorem applies (otherwise, it doesnt). Example Given f(x) = 1/(x 2 – x – 6) and these three intervals: [–3, –1], [–1, 1] and [1, 3], on which interval will the global max/min be guaranteed? For that interval find the global max/min. Extreme Value Theorem Figure 7: When a function is discontinuous on [a, b], global extrema might not exist (as in f(x)) or might exist (as in g(x)). y = f(x) y = g(x) abab global max global min Figure 8: When a function is continuous on (a, b) but not on [a, b], global extrema might not exist (as in f(x)) or might exist (as in g(x)). y = f(x) ab y = g(x) ab global max global min

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Page 29 The Derivative Doesnt Exist? If f(x) is continuous on a closed interval [a, b], the two global extrema will exist. They will occur at c (a < c < b) with either f (c) = 0 or f (c) does not exist, or at a, or at b. In other words, the global extrema will occur at some value(s) between a and b or at the endpoints a and b. The largest of f(a), f(b) and f(c)s will be the global maximum and the smallest of these values will the global minimum. If one thing has not been clear, its the fact that the global max/min can occur at c (a < c < b) with f (c) does not exist. How can that be? What does f (c) does not exist mean? Well, recall that graphically, f (c) means the slope of the tangent line drawn at c if such a tangent line can be drawn, and Extreme Value Theorem the limit definition of f (c) is. So, lets look at the following graphs: Figure 9 y = f(x) (a) c (b) c y = f(x) (c) c y = f(x) (d) c y = f(x) (e) c y = f(x) (f) c y = f(x) (g) c y = f(x) If figure 9(a), f is continuous at c and we can draw a tangent line at c, so f (c) exist (and is equal to whatever the slope of that tangent line is). In 9(b), (c), (d) and (e), f is ______________ at c and we __________ a tangent line at c, so f (c) __________________. In 9(f), f is ______________ at c and the only tangent line can be drawn is a __________ tangent line, but the slope is __________ for a vertical line, so f (c) __________________. In 9(g), f is ______________ at c but a tangent line _____ be drawn, so f (c) __________________. One thing should be obvious is: If f is not continuous at c, then f is not differentiable at c.

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Page 30 Continuity vs. Differentiability Its about time for some true-false questions: 1. If f is not continuous at c, then f is not differentiable at c. (T or F) 2. If f is continuous at c, then f is differentiable at c. (T or F) 3. If f is not differentiable at c, then f is not continuous at c. (T or F) 4. If f is differentiable at c, then f is continuous at c. (T or F) Comments Theorem Applications 1. Show that f(x) = is not differentiable at x = 0, 1 and –2. 2. Show, both graphically and algebraically, that y = |x| + x is continuous at x = 0, but is not differentiable at y = Find the global max and global min of f(x) = |x| – x 2 on [– 1, 2].

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Page 31 Finding the Tangent Line Finding the tangent line? Wait a minute, you might say, we have done this. True, we have done this on page 7 and the box on the right shows how we did it on page 7by using the limit definition to find the slope of the tangent line, i.e., the LONG way, which is totally not needed when we know the SHORTCUT. What shortcut are we talking about? The shortcut of finding derivatives using ____________ ____________ instead of using the tedious limit definition. So, lets redo the same problem using the SHORTCUT. Find an equation of the tangent line of the given function at the given x-value: f(x) = 2x 2 – 3x – 2 at x = 2 Solution: Steps of finding an equation of the tangent line of f(x) at x = x 1 : 1.Start with the point-slope form of a line: y – y 1 = m(x – x 1 ) 2.For y 1 (if its not given), plug x 1 into f, i.e., evaluate f(x 1 ). 3. For m, the slope of the tangent line: i) Find f (x), using derivative formulas of course. ii)Plug x 1 into f, i.e., evaluate f (x 1 ). If it exists, then f (x 1 ) is the slope of the tangent line; if it doesnt, either the tangent line is vertical or the tangent line doesnt exist. More Examples: 1. f(x) = sin x 2 at x = 2.at x = (1, 3)

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