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tangent f(x)=1+4csc(x),\at x=-pi/3

解答

切线

f (x) = 1 + 4 csc (x), at x = - \frac{π}{3}

解答

y = - \frac{8}{3} x + 1 - \frac{8}{3} - \frac{8 π}{9}

求解步骤

找到切点:

(- \frac{π}{3}, 1 - \frac{8}{3})

计算

f (x) = 1 + 4 csc (x)

的斜率:

\frac{df ( x )}{d x} = - 4 cot (x) csc (x)

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{8}{3}

找到斜率 m=

- \frac{8}{3}

且穿过

(- \frac{π}{3}, 1 - \frac{8}{3})

的直线:

y = - \frac{8}{3} x + 1 - \frac{8}{3} - \frac{8 π}{9}

y = - \frac{8}{3} x + 1 - \frac{8}{3} - \frac{8 π}{9}

作图

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