tangent f(x)= 6/(5x+2),(-1,-2)

解答

切线

f (x) = \frac{6}{5 x + 2}, (- 1, - 2)

y = - \frac{10}{3} x - \frac{16}{3}

求解步骤

计算

f (x) = \frac{6}{5 x + 2}

的斜率:

\frac{df ( x )}{d x} = - \frac{30}{( 5 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 = - \frac{10}{3}

找到斜率 m=

- \frac{10}{3}

且穿过

(- 1, - 2)

的直线:

y = - \frac{10}{3} x - \frac{16}{3}

y = - \frac{10}{3} x - \frac{16}{3}

\int (\frac{sin ( 4 x )}{4}) - (\frac{sin ^{3} ( x )}{4}) d x

n = 0 \sum \infty n (n + 1) - 1

\int cos (x + π) d x

in v erse l a pl a ce \frac{e ^{- s}}{s ^{2} ( s + 2 )}

\int 1 + e^{2 x} d x