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

解答

切线

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

解答

y = - \frac{3}{16} x - \frac{1}{2}

求解步骤

找到切点:

(- 2, - \frac{1}{8})

计算

f (x) = \frac{1}{x ^{3}}

的斜率:

\frac{df ( x )}{d x} = - \frac{3}{x ^{4}}

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 = - \frac{3}{16}

找到斜率 m=

- \frac{3}{16}

且穿过

(- 2, - \frac{1}{8})

的直线:

y = - \frac{3}{16} x - \frac{1}{2}

y = - \frac{3}{16} x - \frac{1}{2}

作图

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