Lim x tends to 0 (TaNx SiNx) / X

Lim x tends to 0 (TaNx SiNx) / X

lim x->0 （tanx-sinx)/x
=lim x->0 （1/cosx-1)sinx/x
=lim x->0 (1/cosx-1)
=1-1
=0
It is known that the function f (x) = ax + B is an increasing function, and its domain and range are [1,2]
(1) Finding the analytic expression of F (x)
(2) If G (x) = f (x) (x ≥ 1) g (x) = g (x + 2) (x < 1), find g (- 2) and G (the value of log0.2 bottom 0.3)
If G (x) = f (x) (x ≥ 1), G (x) = g (x + 2) (x < 1), then G (- 2) = g (- 2 + 2) = g (0 + 2) = f (2) = 2
We know that the domain of F (x) = x ^ 2-ax + 3 is [2,4]. When a belongs to [2,6], we can find its range
1. When a belongs to [2,6], find its range
2. If the maximum value of F (x) is - 9, find the value of A
F (x) = (x-a / 2) ^ 2 + 3-A ^ 2 / 4, axis of symmetry x = A / 2
1) Because 2
The tangent of the image of the function f (x) = x & # 179; + ax & # 178; + B at point P (1,0) is parallel to the straight line 3x + y = 0 (1)
The tangent of function f (x) = x & # 179; + ax & # 178; + B at point P (1,0) is parallel to the straight line 3x + y = 0 (1): find a, B (2): find the maximum value of function f (x) in ＜ 0, t ＞ (t ＞ 0)
(1) F (x) is defined as X ∈ R. f '(x) = 3x & # 178; + 2aX, f' (1) = 3 + 2A = - 3, so a = - 3f (1) = 1-3 + B = 0, so B = 2, so a = - 3, B = 2. (2) f (x) = x & # 179; - 3x & # 178; + 2, f '(x) = 3x & # 178; - 6x when f' (x) = 0, 3x & # 178; - 6x = 0, that is, X (X-2) = 0, the solution is X1 = 0, X2 = 2
It is known that a is a set composed of nonnegative even numbers less than 5, and B is a set composed of 0, x, X & # 178;; if the elements of the two sets are the same, the value of X is obtained
analysis
Element 0 2 4 of set a
B（0 x x²)
When x = 2
Set B (0 2 4) satisfies the problem
When x = 4 is
Set B (0 4 16)
A and B are not equal
So it's against the theme
So x = 2
Given the function FX = x2 + 2aX + 3, when a is equal to - 1, find the monotone increasing interval of FX, if a is equal to negative one and X belongs to [- 1.2], find the maximum of FX
f'(x)=2x-2;
Let f '(x) = 0, then x = 1;
f(1)=2;
f(-1)=6
f(2)=3;
So the maximum is 6
The set a composed of positive integers satisfies the condition that if a ∈ a, then 12 / a ∈ a,
Since a and 12 / a belong to a, and all elements of a are positive integers, all elements of a should be factors of 12, that is, 1,2,3,4,6,12. And (1,12), (3,4), (2,6) must appear in pairs. Considering that there are four elements of a, a can have the following three cases: (1) a = {1,3,4,12} (2) a = {1,2,6
For any differentiable function f (x) on R, if (x-1) f '(x) > = 0, then a.f (0) + F (2) 2F (1)
When x ≥ 1, f '(x) ≥ 0, then f (x) is an increasing function or a constant function,
∴F(2)≥F(1),
When x ≤ 1, f '(x) ≥ 0,
Then f (x) is a decreasing function or a constant function,
∴F(0)≥F(1),∴F(0)+F(2)≥2F(1).
The set of positive integers greater than 7 is () a {1,2,3,4,5,6,7} B {x | x ≤ 7}
Is 0 not a positive integer? What does x | in front of B option x ≤ 7 mean
Positive, 0, negative: This is the classification of real numbers
{x | x ≤ 7}, which is a kind of set representation, means: the value is x, and X is all numbers ≤ 7
F (x) + F (y) = f (x + y) + 2 x ＞ 2F (x ＞ 2) is proved to be an increasing function
F (x) + F (y) = f (x + y) + 2 x ＞ 2F (x ＞ 2) is proved to be an increasing function and f (A & # 178; - 2a-2) ＜ 3
It's x ＞ 2 and f (x) ＞ 2
Let X1 > x2 > 2
Then f (x1) - f (x2) = f (x1-x2 + x2) - f (x2) = f (x1-x2) + F (x2) - 2-F (x2) = f (x1-x2) - 2 > 0
That is, f (x1) > F (x2)
Therefore, it is an increase
Do it yourself!
Correct the mistake! When x > 0, f (x) > 2. In addition, the answer should be f (A & # 178; - 2a-2) < 3.
I don't know how the landlord's attitude towards learning mathematics is. If he is so careless and careless, he will definitely not be able to learn well
High school math. It seems that we must first correct our attitude towards learning!
High school mathematics is very rigorous, many topics. I mean it. ... unfold
Correct the mistake! When x > 0, f (x) > 2. In addition, the answer should be f (A & # 178; - 2a-2) < 3.
I don't know how the landlord's attitude towards learning mathematics is. If he is so careless and careless, he will definitely not be able to learn well
High school math. It seems that we must first correct our attitude towards learning!
High school mathematics is very rigorous, many topics. I mean it. Put it away
Is it necessary to prove that f (x) is an increasing function or that f (x) > 2?