tangent f(x)=6x^2-4x+4,(1,6)

解答

切线

f (x) = 6 x^{2} - 4 x + 4, (1, 6)

y = 8 x - 2

求解步骤

计算

f (x) = 6 x^{2} - 4 x + 4

的斜率:

\frac{df ( x )}{d x} = 12 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 = 8

找到斜率 m=

8

且穿过

(1, 6)

的直线:

y = 8 x - 2

y = 8 x - 2

d er i v a t i v e y = \frac{3}{e ^{x} + e ^{- x}}

\frac{d}{d x} (3 x, - 2 y, x - y)

\frac{d}{d x} ((e^{- x^{2}}) (- 2 x))

y^{^{''}} + y = - sin (2 x), y (0) = 0, y^{^{'}} (0) = 0

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