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tangent y=(3sqrt(x)+x)(2-x^2)

解答

切线

y = (3 x + x) (2 - x^{2})

解答

y = (- 3 a_{0}^{2} - 6 a_{0} a_{0} - \frac{3 a _{0}^{\frac{3}{2}}}{2} + \frac{3}{a _{0}} + 2) x + 2 a_{0}^{3} + 3 a_{0}^{2} a_{0} + 3 a_{0} + \frac{3 a _{0}^{\frac{5}{2}}}{2}

求解步骤

计算在一般点

x = a_{0}

的切线

找到切点:

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

计算

y = (3 x + x) (2 - x^{2})

的斜率:

\frac{d y}{d x} = - 3 x^{2} - 6 x x - \frac{3 x ^{\frac{3}{2}}}{2} + \frac{3}{x} + 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 = - 3 a_{0}^{2} - 6 a_{0} a_{0} - \frac{3 a _{0}^{\frac{3}{2}}}{2} + \frac{3}{a _{0}} + 2

找到斜率 m=

- 3 a_{0}^{2} - 6 a_{0} a_{0} - \frac{3 a _{0}^{\frac{3}{2}}}{2} + \frac{3}{a _{0}} + 2

且穿过

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

的直线:

y = (- 3 a_{0}^{2} - 6 a_{0} a_{0} - \frac{3 a _{0}^{\frac{3}{2}}}{2} + \frac{3}{a _{0}} + 2) x + 2 a_{0}^{3} + 3 a_{0}^{2} a_{0} + 3 a_{0} + \frac{3 a _{0}^{\frac{5}{2}}}{2}

y = (- 3 a_{0}^{2} - 6 a_{0} a_{0} - \frac{3 a _{0}^{\frac{3}{2}}}{2} + \frac{3}{a _{0}} + 2) x + 2 a_{0}^{3} + 3 a_{0}^{2} a_{0} + 3 a_{0} + \frac{3 a _{0}^{\frac{5}{2}}}{2}

作图

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