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tangent f(x)=2x^3+42x^2+288x+14

解答

切线

f (x) = 2 x^{3} + 42 x^{2} + 288 x + 14

解答

y = (6 a_{0}^{2} + 84 a_{0} + 288) x - 4 a_{0}^{3} - 42 a_{0}^{2} + 14

求解步骤

计算在一般点

x = a_{0}

的切线

找到切点:

(a_{0}, 2 a_{0}^{3} + 42 a_{0}^{2} + 288 a_{0} + 14)

计算

f (x) = 2 x^{3} + 42 x^{2} + 288 x + 14

的斜率:

\frac{df ( x )}{d x} = 6 x^{2} + 84 x + 288

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 = 6 a_{0}^{2} + 84 a_{0} + 288

找到斜率 m=

6 a_{0}^{2} + 84 a_{0} + 288

且穿过

(a_{0}, 2 a_{0}^{3} + 42 a_{0}^{2} + 288 a_{0} + 14)

的直线:

y = (6 a_{0}^{2} + 84 a_{0} + 288) x - 4 a_{0}^{3} - 42 a_{0}^{2} + 14

y = (6 a_{0}^{2} + 84 a_{0} + 288) x - 4 a_{0}^{3} - 42 a_{0}^{2} + 14

作图

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