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](/uploads/image/z/7693099-43-9.jpg?t=%E8%8B%B1%E8%AF%AD%E7%BF%BB%E8%AF%91Prove+that+the+curves+y%3Dx%5E3+and+y%3D%28x%2B1%29%2F%28x%5E2+%2B4%29+have+exactly+one+point+in+common%2Cand+use+differentiation+to+find+the+gradient+of+each+curve+at+this+point.%E9%A2%98%E7%9B%AE%E5%A4%A7%E6%84%8F%E5%A4%A7%E6%A6%82%E6%98%AF%E8%BF%99%E6%A0%B7%EF%BC%9A%E6%B1%82%E8%AF%81%E5%BC%A7+y%3Dx%5E3+%E5%92%8C%E5%BC%A7+y%3D%28x%2B1%29%2F%28x%5E2+%2B4%29+%E6%81%B0%E6%9C%89%E4%B8%80%E4%B8%AA)
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。