When finding the limit of high number, when x → 0, it is equivalent to infinitesimal When finding the limit, if x → 0, for example, SiNx appears in any form, (its square, plus or minus other coefficient factors), can SiNx in the factor be replaced by X? Or only when it can be multiplied by F (x) = sinxg (x)? (it's easy to find the answer when changing to x)

When finding the limit of high number, when x → 0, it is equivalent to infinitesimal When finding the limit, if x → 0, for example, SiNx appears in any form, (its square, plus or minus other coefficient factors), can SiNx in the factor be replaced by X? Or only when it can be multiplied by F (x) = sinxg (x)? (it's easy to find the answer when changing to x)

Equivalent infinitesimals cannot be replaced in addition and subtraction, but this is not absolute. When two infinitesimals are subtracted, if they are not equivalent infinitesimals, they can be replaced by their equivalent infinitesimals respectively. Similarly, if two infinitesimals are added, their corresponding limit is not equal to - 1, they can be replaced by their equivalent infinitesimals respectively

Given that f (x) = x ^ (3) - ax is a monotonically increasing function on [1, positive infinity), what is the maximum value of a?

F (x) = x ^ (3) - ax, then f (x) derivative = 3 * x ^ 2-A, the derivative increases from 0 to positive infinity and decreases from negative infinity to 0, so the derivative increases at [1, positive infinity), so the minimum value of F (x) derivative is 3 * 1-A greater than or equal to 0, that is, a less than or equal to 3, so the maximum value is 3

Given that the tangent of the function y = f (x) at point (2,1) is parallel to the straight line 3x-y-2 = 0, then y ′| x = 2 is equal to () A. -3 B. -1 C. 3 D. 1

The tangent slope of the function y = f (x) at point (2,1) is equal to the derivative of the function at point (2,1)
∵ the tangent of function y = f (x) at point (2, 1) is parallel to the straight line 3x-y-2 = 0,
∴y′|x=2=3.
Therefore: C

It is known that the tangent of the image of the function f (x) = MX3 + NX2 at point (- 1,2) is exactly parallel to the straight line 3x + y = 0. If f (x) monotonically decreases in the interval [T, t + 1], the value range of the real number t is ___

From the known conditions, f '(x) = 3mx2 + 2nx,
From F '(- 1) = - 3, ‡ 3m-2n = - 3
And f (- 1) = 2, ‡ - M + n = 2,
∴m=1,n=3
∴f(x)=x3+3x2,∴f'(x)=3x2+6x.
Let f '(x) < 0, that is, X2 + 2x < 0,
The monotone decreasing interval of function f (x) is (- 2, 0)
∵ f (x) monotonically decreases on the interval [T, t + 1],
Then the value range of real number T is [- 2, - 1]
So the answer is [- 2, - 1]

It is known that integers a, B, C and D satisfy ABCD = 25, and a > b > C > D, then | a + B | + | C + D | is equal to () A. O B. 10 C. 2 D. 12

25=5 × five × one × 1=5 × (-5) × one × (-1),
Then a = 5, B = 1, C = - 1, d = - 5,
∴|a+b|+|c+d|=|5+1|+|-1-5|
=6+6=12.
Therefore, D

If 1-A = absolute value of A-2007 + √ 2007-a, find 2006 ²+ a

2007-a ≥ 0A ≤ 2007, so 1-A = 2007-a + √ 2007-a √ 2007-a = - 20062007-a = 2006 ² two thousand and six ²+ A = 2007 click [evaluation] in the upper right corner, and then you can select [satisfied, the problem has been perfectly solved]. If there are new problems, please do not send them in the form of questioning, and send them to