標題:

Amath微分劃圖問題

 

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發問:

Sketch the curve y = (x+1)/(x^2+3) 可否給我祥細步驟。

最佳解答:

y = (x+1)/(x^2+3) dy/dx = (x^2+3)/(x^2+3)^2 - 2x(x+1)/(x^2+3)^2 = (-x^2-2x+3)/(x^2+3)^2 = -(x-1)(x+3)/(x^2+3)^2 d^2y/dx^2 = (-2x-2)/(x^2+3)^2 - 2(2x)(-x^2-2x+3)/(x^2+3)^3 = (-2x-2)/(x^2+3)^2 - 4x(-x^2-2x+3)/(x^2+3)^3 = [-2(x+1)(x^2+3) + 4x(x^2+2x-3)]/(x^2+3)^3 = (-2x^3-2x^2-6x-6+4x^3+8x^2-12x)/(x^2+3)^3 = (2x^3+6x^2-18x-6)/(x^2+3)^3 = 2(x^3+3x^2-9x-3)/(x^2+3)^3 When x = 0, y = 1/3 -> y-intercept = 1/3 When x = -1, y = 0 -> x-intercept = 0 When x = 1, dy/dx = 0 and d^2y/dx^2 < 0 -> (1,1/2) is the local maximum When x = -3, dy/dx = 0 and d^2y/dx^2 > 0 -> (-3,-1/6) is the local minimum lim(x->inf) y = lim(x->inf) (x+1)/(x^2+3) = lim(x->inf) (1+1/x)/(x+3/x) = 0 -> y = 0 is a horizontal asymptote Since x^2+3 >= 3 > 0 for any real x, there are no vertical asymptotes Optional: When x ≈ -4.76, -0.305 or 2.06, d^2y/dx^2 ≈ 0 -> (-4.76,-0.147), (-0.305,0.225), (2.06,0.422) are the points of inflection The required curve is sketched at https://www.dropbox.com/s/ursp4yr7duev2st/7013052600070.PNG

其他解答:13413D6BADCE0D57
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