標題:
derived functions
發問:
Please do no.1, no. 2 and no .7 that show on the following link. http://cid-632ec8cb72303e4e.skydrive.live.com/self.aspx/ex/%e6%8e%83%e6%8f%8f0012.jpg 更新: u must solve the problems by using derived function 更新 2: 1. Find the slope of the tangent to the curve at the point (1,1) 更新 3: no equation provided in Ques no.1 更新 4: maybe Ques 1 has some problems, you can do the other two.
最佳解答:
Your first question is not clear. The equation of the curve cannot be read. 2009-12-12 00:16:19 補充： (2)(a) y = 4x^2 - 2x - 1 dy/dx = 8x - 2 Gradient at (1,1) is 8 - 2 = 6 (b) dy/dx = 0 => x = 1/4 d^2y/dx^2 = 8 => min Minimum of y = 4*(1/4)^2 - 2(1/4) - 1 = 0.25 - 0.5 - 1 = -1.25 (c) The curve cuts the x-axis when y = 0 0 = 4x^2 - 2x - 1 x = [2 +/- √(4 + 16)]/8 x = 1/4 +/- (√5)/4 = -0.309 or 0.809 The curve cuts the x-axis at (-0.309,0) and (0.809, 0) (7)(i) s = t^2 - 4t velocity = ds/dt = 2t - 4 at t = 0, velocity = -4 (ii) velocity at t = 2 is 2(2) - 4 = 0 (iii) Displacement in first two seconds = 2^2 - 4*2 = -4 Average velocity = -4/2 = -2 (iv) when the point is at rest, the velocity = 0 2t - 4 = 0 => t = 2 s(2) = 2^2 - 4*2 = - 2
derived functions
發問:
Please do no.1, no. 2 and no .7 that show on the following link. http://cid-632ec8cb72303e4e.skydrive.live.com/self.aspx/ex/%e6%8e%83%e6%8f%8f0012.jpg 更新: u must solve the problems by using derived function 更新 2: 1. Find the slope of the tangent to the curve at the point (1,1) 更新 3: no equation provided in Ques no.1 更新 4: maybe Ques 1 has some problems, you can do the other two.
最佳解答:
Your first question is not clear. The equation of the curve cannot be read. 2009-12-12 00:16:19 補充： (2)(a) y = 4x^2 - 2x - 1 dy/dx = 8x - 2 Gradient at (1,1) is 8 - 2 = 6 (b) dy/dx = 0 => x = 1/4 d^2y/dx^2 = 8 => min Minimum of y = 4*(1/4)^2 - 2(1/4) - 1 = 0.25 - 0.5 - 1 = -1.25 (c) The curve cuts the x-axis when y = 0 0 = 4x^2 - 2x - 1 x = [2 +/- √(4 + 16)]/8 x = 1/4 +/- (√5)/4 = -0.309 or 0.809 The curve cuts the x-axis at (-0.309,0) and (0.809, 0) (7)(i) s = t^2 - 4t velocity = ds/dt = 2t - 4 at t = 0, velocity = -4 (ii) velocity at t = 2 is 2(2) - 4 = 0 (iii) Displacement in first two seconds = 2^2 - 4*2 = -4 Average velocity = -4/2 = -2 (iv) when the point is at rest, the velocity = 0 2t - 4 = 0 => t = 2 s(2) = 2^2 - 4*2 = - 2
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