Pre-Calc Homework Solutions 451

# Pre-Calc Homework Solutions 451 - Appendix A2 dx 0 at d 2 4...

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Unformatted text preview: Appendix A2 dx 0 at d 2 4 dx dy and at , and vertical tangents 0, 0 3 3 d d 5 at , at , and at . 3 3 dy 0 For 0 (or 2 ), becomes and l'Hpital's Rule dx 0 dy d 451 There are horizontal tangents 0, s Appendix A2 (pp. 581584) 1. Step 1: The formula holds for n ...x x Step 2: Suppose x 1 2 1, because x1 x1 1 Then, x1 x2 x1 ... x2 k x2 xk ... x1 . xk . xk ... xk 1 by the leads to dy dx 0 triangle inequality. sin (0) 4 sin (0) cos (0) cos (0) 2 cos 2 (0) 2 sin 2 (0) 0, so this is So, by the transitivity of , x1 x1 x2 ... xk 1 original formula for all n. x2 ... xk 1 . another horizontal tangent line. Horizontal tangents: At 0 or 2 , we have r 0 and the Cartesian 0. The mathematical induction principle now guarantees the 2. Step 1: The formula holds for n 1 r (1 r)(1 1 r r) 1 1 1, because r2 . r coordinates are (0, 0), so the tangent is y At are 2 , we have r 3 3 2 3 2 cos , sin 2 3 2 3 3 and the Cartesian coordinates 2 3 3 3 , , so the tangent is 4 4 Step 2: Suppose 1 1 r 1 r ... 1 r2 rk r k 1(1 r ... 1 rk 1 1 rk 1 . Then 1 r r2 rk 3 3 . y 4 4 , we again have r At 3 1 3 and the Cartesian 2 3 , 4 3 4 3 rk 1 rk 1 1 r r) 1 rk 2 . 1 r The mathematical induction principle now guarantees the original formula for every positive integer n. , so 3. Step 1: The formula holds for n Step 2: Suppose d k (x ) dx d k 1 d (x ) (x dx dx coordinates are the tangent is y 3 4 3 4 cos , sin 2 3 2 3 3 4 3 1, because d (x) dx 1. . kx k 1. Then x k) xk x kx k d k (x ) dx xk (k Vertical tangents: At are 1 1 cos , sin 2 3 2 3 1 3 , , so the tangent is x 4 4 1 . 4 3 d (x) dx x kx k 1 and the Cartesian coordinates 2 1 xk 1)x k. , we have r The mathematical induction principle now guarantees the original formula for any positive integer n. 4. Step 1: The formula holds for n 1, because f (x1) f (x1). Step 2: Suppose f (x1x2 ... xk) f (x1) f (x2) ... f (xk). Then by the given property, f (x1x2 ... xk 1) f (x1x2 ... xk) f (xk 1) f (x1) f (x2) ... f (xk 1). The mathematical induction principle now guarantees the original formula for every positive integer n. 5. Step 1: The formula holds for n Step 2: Suppose 2 31 2 32 2 31 2 32 At , we have r 2 and the Cartesian coordinates are ( 2, 0), so the tangent is x 2. (2 cos , 2 sin ) At are x y x 5 , we have r 3 1 5 1 5 cos , sin 2 3 2 3 1 and the Cartesian coordinates 2 1 , 4 3 4 , so the tangent is 0, 1 . In summary, the horizontal tangents are y 4 3 4 3 1, because 2 3k 2 3 1 1 . 3 , and y 1 . 4 3 4 3 , and the vertical tangents are ... 2 3k 1 1 1 3k ... 3 2 3k 1 1 1 3k 1. 1 . Then 3k 2 3k 1 2 and x 1 1 The mathematical induction principle now guarantees the original formula for all positive integers n. ...
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## This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.

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