Pre-Calc Homework Solutions 167

# Pre-Calc Homework Solutions 167 - Section 4.4(c Using our...

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Unformatted text preview: Section 4.4 (c) Using our expression for the y-intercept of the tangent line, the height of the triangle is a mx f (x) 5 5 1 2 1 2 167 f (x) x 100 100 x2 x2 x 2 100 x2 2 100 x2 x2 x We may use graphing methods or the analytic method in part (d) to find that the minimum value of A(x) occurs at x 8.66. Substituting this value into the expression above, the height of the triangle is 15. This is 3 times the y-coordinate of the center of the ellipse. (d) Part (a) remains unchanged. The domain is (0, C). To graph, note that f (x) f (x) B B C 2 B 1 1 C2 x2 C2 x2 B B C C2 Bx x 2 and x2 ( 2x) C C2 . Therefore, we have f (x) 2 f (x) Bx C C2 x2 B B C C2 Bx x2 x2 2 A(x) f (x) x x C C2 Bx C C2 1 BCx C2 1 BCx C2 1 BCx BC(C x C2 C2 x2 x (BC B C2 x2) Bx C2 x2 2 x2 [Bx 2 [Bx 2 (BC B C2 x2 x 2)]2 x 2)( B(C 2 C2 x 2)]2 x2 BC C 2 C2 x 2)]2 x2 C2 [BC(C x 2)2 x2 (x C 2 x 2)(2)(C C2 x 2) x C2 x2 (C x 2) C2 x 2 )2 x x C2 x2 C2 x 2(1) A (x) BC x 2(C 2 C 2 x 2) BC(C 2 2 x (C x 2) 2x 2 (C C2 x 2) ( x2 C 2 x2 C2 x2 ) ] C 2 x 2) BC(C 2 2 x (C x 2) [ ( 2x 2 Cx2 C2 x2 C C2 x2 x2 (C 2 x 2) BC(C C 2 x 2) 2 2 x (C x 2) Cx 2 C2 x2 C C2 x2 C2 ) x 2] BC(C x 2(C 2 BC 2(C x 2(C 2 C 2 x 2) [Cx 2 x 2)3/2 C 2 x 2) (2x 2 x 2)3/2 C(C 2 x 2) C2 C2 C2 C C2 x 2) ...
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