英语翻译Prove that the curves y=x^3 and y=(x+1)/(x^2 +4) have ex
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英语翻译
Prove that the curves y=x^3 and y=(x+1)/(x^2 +4) have exactly one point in common,and use differentiation to find the gradient of each curve at this point.
题目大意大概是这样:
求证弧 y=x^3 和弧 y=(x+1)/(x^2 +4) 恰有一个公共点,并用微分法求出两道弧与该点的斜率.
Prove that the curves y=x^3 and y=(x+1)/(x^2 +4) have exactly one point in common,and use differentiation to find the gradient of each curve at this point.
题目大意大概是这样:
求证弧 y=x^3 和弧 y=(x+1)/(x^2 +4) 恰有一个公共点,并用微分法求出两道弧与该点的斜率.
y = x³ = (x + 1)(x² + 4) = x³ + x² + 4x + 4
x² + 4x + 4 = (x + 2)² = 0
x = -2
The common point is (-2, -8)
y = x³, y' = 2x²; x = -2, y' =2*(-2)² = 8
y = (x + 1)(x² + 4)
y' = x² + 4 + (x + 1)*2x = 3x² + 2x + 4
x = -2, y' = 3(-2)² +2(-2) + 4 = 12
再问: 是(x + 1)/(x² + 4)哦,中间有个除……
再答: Sorry I missed that. It would make it a lot more difficult. I'll take another look when I have time.
x² + 4x + 4 = (x + 2)² = 0
x = -2
The common point is (-2, -8)
y = x³, y' = 2x²; x = -2, y' =2*(-2)² = 8
y = (x + 1)(x² + 4)
y' = x² + 4 + (x + 1)*2x = 3x² + 2x + 4
x = -2, y' = 3(-2)² +2(-2) + 4 = 12
再问: 是(x + 1)/(x² + 4)哦,中间有个除……
再答: Sorry I missed that. It would make it a lot more difficult. I'll take another look when I have time.
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