tangent f(x)=(x-4)(x^2+8),(1,-27)

解答

切线

f (x) = (x - 4) (x^{2} + 8), (1, - 27)

y = 3 x - 30

求解步骤

计算

f (x) = (x - 4) (x^{2} + 8)

的斜率:

\frac{df ( x )}{d x} = 3 x^{2} - 8 x + 8

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 = 3

找到斜率 m=

3

且穿过

(1, - 27)

的直线:

y = 3 x - 30

y = 3 x - 30

x \to 1 - lim (- x^{2} + 5 x + (- 4))

n = 3 \sum \infty \frac{2}{n ln ^{2} ( n )}

a re a y = x, y = 3 x

\int \frac{x}{( 5 x ^{2} + 2 ) ^{6}} d x

\int_{0}^{1} \frac{1}{1 + x} d x