Find the point on the line 2x - 2y = 9 that is closest to the point ((0, 1) (Che

Question

Find the point on the line 2x - 2y = 9 that is closest to the point ((0, 1) (Check your answer with the grapher) Find the general antiderivative of (a) f(x) = 0, (b) f(x) - x^-1/3 + 1, (c) f(x) = e^x + squareroot 3x, (d) f(x) = sin x + 3 cos x. State the Evaluation Part of the Fundamental Theorem of Calculus. Then find the definite integrals: (a)f^4_2 t^-3 dt. (b) Find integral^0_-2(e^x + x) dx. (c) Find integral^p_2 tdt (the answer is a function of p). State the Differentiation Part of the Fundamental Theorem of Calculus. Then find a d/dx integral^x_2 cos(t^4) dt. b Find d/dx integral^6_x cos (squareroot s^4 + 1)ds. C Find d/dx integral^2x + 1_2 In(t + 1)dt. d Find d/dx integral^x_-x z + 1/z + 2 dz. e Find d/dx integral^2_-3x 2^t2 dt.

Explanation / Answer

(16) Let the point on the line be (h, k)

Then 2h - 2k = 9

2k = 2h - 9

k = (2h - 9)/2

Distance between (h, k) and (0, 1) is D = {h^2 + (k - 1)^2}

Substituting for k as (2h - 9)/2 and simplifying, we get D = (8h^2 + 121 - 44h)/4}

For minimum D, dD/dh = 0

(4h - 11) / [2 {(8h^2 + 121 - 44h)/4}] = 0

4h - 11 = 0

h = 11/4 and k = [2(11/4) - 9]/2 = -7/4

The point is (h, k) = (11/4, -7/4)

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