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tangent f(x)=x^3+2x,\at x=-1

解答

切线

f (x) = x^{3} + 2 x, at x = - 1

解答

y = 5 x + 2

求解步骤

找到切点:

(- 1, - 3)

计算

f (x) = x^{3} + 2 x

的斜率:

\frac{df ( x )}{d x} = 3 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 = 5

找到斜率 m=

5

且穿过

(- 1, - 3)

的直线:

y = 5 x + 2

y = 5 x + 2

作图

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