Unformatted text preview: f ( x ) = sin x + cos x . The derivative is f ± ( x ) = cos xsin 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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 Spring '07
 JonathanRogawski
 Math, Calculus, Critical Point, Mathematical analysis, critical value, local maximum

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