tangent f(x)=x^2+8,(-2,12)

解答

切线

f (x) = x^{2} + 8, (- 2, 12)

y = - 4 x + 4

求解步骤

计算

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

的斜率:

\frac{df ( x )}{d x} = 2 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 = - 4

找到斜率 m=

- 4

且穿过

(- 2, 12)

的直线:

y = - 4 x + 4

y = - 4 x + 4

\int (x^{2} + 1)^{9} d x

\int x^{2} x^{2} - 1 d x

\int x (3 x + 1)^{2} d x

x \to 1 lim (ln (\frac{1}{e} x))

\frac{d ^{2}}{d x ^{2}} (5 (x^{2} + 8)^{4})