We know that the function f (x) = 3 / 2x to the power of 3-2x-6x + 1 Given the function f (x) = the cubic power of 2x-2x-6x + 1 (1) Find the tangent equation of the curve f (x) at x = 0; (2) Find the maximum and minimum value of function f (x) in the interval [- 3,1],

We know that the function f (x) = 3 / 2x to the power of 3-2x-6x + 1 Given the function f (x) = the cubic power of 2x-2x-6x + 1 (1) Find the tangent equation of the curve f (x) at x = 0; (2) Find the maximum and minimum value of function f (x) in the interval [- 3,1],


F '(x) = 2x & # 178; - 4x-6 = 2 (x-3) (x + 1) 1, f' (0) = - 6, that is, the tangent slope k of the curve at x = 0 = - 6, the tangent point is (0,1), then the tangent equation is: 6x + Y-1 = 0; 2, f '(x) increases on [- 3, - 1], decreases on [- 1,1], and f (- 3) = - 17, f (- 1) = 13 / 3, f (1) = - 19 / 3, then



If the function f (x) = A & nbsp; x2 + 2x − 3 + m (a > 1) passes through the point (1,10), then M=______ .


When x = 1, f (x) = A0 + M = 10, M = 10-1 = 9



If f (x) = the third power of x-x, the second power of X-1, then f (f (1))=


f(x)=x^3-x^2-1
f(1)=1^3-1^2-1=-1
f(f(1))=(-1)^3-(-1)^2-1=-1-1-1=-3



Given the cubic power of the function FX = 24x-2x (1), find the monotone interval of the function (2). When x ∈ [- 4,3], find the maximum and minimum of the function


f'=24-6x²=-6(x+2)(x-2)=0
X = - 2, or x = 2
f'>0
-2



The calculation result of 20082008 × 2007-20072007 × 2008 is ()
A. 0B. 2007C. 2008


20082008 × 2007-20072007 × 2008 = (20080000 + 2008) × 2007 - (20070000 + 2007) × 2008 = 20080000 × 2007 + 2008 × 2007 - (20070000 × 2008 + 2007 × 2008) = 20080000 × 2007 + 2008 × 2007-20070000 × 2008-2007 × 2008 = 0



As shown in the figure, ⊙ o diameter CD = 10, chord AB = 8, ab ⊥ CD, perpendicular foot m, find the length of DM


The diameter of connecting OA, ∵ o is CD = 10, ∵ OA = 5, ∵ chord AB = 8, ab ⊥ CD, ∵ am = 12ab = 12 × 8 = 4. In RT △ AOM, OM = oa2 − AM2 = 52 − 42 = 3, ∵ DM = od + om = 5 + 3 = 8



12 × 3.14 + 62 × 3.14 + 17 × 3.14 by factorization


Original formula = 3.14 (12 + 62 + 17)
=3.14x91
=3.14(100-9)
=314-28.26
=285.74



As shown in the figure, in △ ABC, ∠ BAC = 120 ° ad ⊥ BC is in D, and ab + BD = DC, then ∠ C=______ Degree


Cut off de = dB on DC, connect AE, let ∠ C = x, ∵ AB + BD = DC, de = dB, ∵ CE = AB, and ∵ ad ⊥ BC, DB = De, ∵ straight line ad is the vertical bisector of be, ∵ AB = AE, ∵ CE = AE, ∵ B = ∠ AEB, ∵ C = ∠ CAE, and