作业帮英语翻译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
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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
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.
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
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
微分法？是微积分么？求导？楼主你是什么学校的？相信的数学和英语一定比我好，我这个菜鸟就不关公面前舞大刀了，相信楼主一定可以的。你加油吧。就、就是求导……我海外的……在零点斜率相等，求导，联立，求出公共点的式子，再来分别根据式子分类画个图求值。看不懂求详细…………...
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微分法？是微积分么？求导？楼主你是什么学校的？相信的数学和英语一定比我好，我这个菜鸟就不关公面前舞大刀了，相信楼主一定可以的。你加油吧。
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求解函数F(x)=x^3-(x+1)/(x^2+4)只有一个零点。求出零点的x值，并求y=x^3的导数在该点的值。