121616949-math.110

# 121616949-math.110 - f x = sin x cos x The derivative is f...

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96 Chapter 5 Curve Sketching close together. Nevertheless, because this method is conceptually simple and sometimes easy to perform, you should always consider it. EXAMPLE 5.2 Find all local maximum and minimum points for the function f ( x ) = x 3 - x . The derivative is f ( x ) = 3 x 2 - 1. This is defined everywhere and is zero at x = ± 3 / 3. Looking first at x = 3 / 3, we see that f ( 3 / 3) = - 2 3 / 9. Now we test two points on either side of x = 3 / 3, making sure that neither is farther away than the nearest critical value; since 3 < 3, 3 / 3 < 1 and we can use x = 0 and x = 1. Since f (0) = 0 > - 2 3 / 9 and f (1) = 0 > - 2 3 / 9, there must be a local minimum at x = 3 / 3. For x = - 3 / 3, we see that f ( 3 / 3) = 2 3 / 9. This time we can use x = 0 and x = - 1, and we find that f ( - 1) = f (0) = 0 < 2 3 / 9, so there must be a local maximum at x = - 3 / 3. Of course this example is made very simple by our choice of points to test, namely x = - 1, 0, 1. We could have used other values, say - 5 / 4, 1 / 3, and 3 / 4, but this would have made the calculations considerably more tedious. EXAMPLE 5.3 Find all local maximum and minimum points for
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Unformatted text preview: f ( x ) = sin x + cos x . The derivative is f ± ( x ) = cos x-sin x . This is always deﬁned and is zero whenever cos x = sin x . Recalling that the cos x and sin x are the x and y coordinates of points on a unit circle, we see that cos x = sin x when x is π/ 4, π/ 4 ± π , π/ 4 ± 2 π , π/ 4 ± 3 π , etc. Since both sine and cosine have a period of 2 π , we need only determine the status of x = π/ 4 and x = 5 π/ 4. We can use 0 and π/ 2 to test the critical value x = π/ 4. We ﬁnd that f ( π/ 4) = √ 2, f (0) = 1 < √ 2 and f ( π/ 2) = 1, so there is a local maximum when x = π/ 4 and also when x = π/ 4 ± 2 π , π/ 4 ± 4 π , etc. We use π and 2 π to test the critical value x = 5 π/ 4. The relevant values are f (5 π/ 4) =-√ 2, f ( π ) =-1 >-√ 2, f (2 π ) = 1 >-√ 2, so there is a local minimum at x = 5 π/ 4, 5 π/ 4 ± 2 π , 5 π/ 4 ± 4 π , etc....
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