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tangent y=x^4+3x^3-2x-2

解答

切线

y = x^{4} + 3 x^{3} - 2 x - 2

解答

y = (4 a_{0}^{3} + 9 a_{0}^{2} - 2) x - 3 a_{0}^{4} - 6 a_{0}^{3} - 2

求解步骤

计算在一般点

x = a_{0}

的切线

找到切点:

(a_{0}, a_{0}^{4} + 3 a_{0}^{3} - 2 a_{0} - 2)

计算

y = x^{4} + 3 x^{3} - 2 x - 2

的斜率:

\frac{d y}{d x} = 4 x^{3} + 9 x^{2} - 2

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 4 a_{0}^{3} + 9 a_{0}^{2} - 2

找到斜率 m=

4 a_{0}^{3} + 9 a_{0}^{2} - 2

且穿过

(a_{0}, a_{0}^{4} + 3 a_{0}^{3} - 2 a_{0} - 2)

的直线:

y = (4 a_{0}^{3} + 9 a_{0}^{2} - 2) x - 3 a_{0}^{4} - 6 a_{0}^{3} - 2

y = (4 a_{0}^{3} + 9 a_{0}^{2} - 2) x - 3 a_{0}^{4} - 6 a_{0}^{3} - 2

作图

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