Given the function FX = 3sin (a 2x + Wu / 3), find the monotone decreasing interval of the function

Given the function FX = 3sin (a 2x + Wu / 3), find the monotone decreasing interval of the function

If the function y = 3x & sup2; + 2 (A-1) x + 1 is a decreasing function in the interval (- ∞, 1), find the value range of A

Because in (- ∞, 1) is reduced, so - B / 2A ≥ 1, take the number to solve the equation

Monotone decreasing interval of function y = 2 ^ (4 + 3x-x ^ 2)

Because the exponent is 2, greater than 1
So the monotone decreasing interval of Y is the monotone decreasing interval of G (x) = 4 + 3x-x ^ 2
The opening of G (x) is downward, and the axis of symmetry is x = 3 / 2
When x > 3 / 2, G (x) decreases monotonically
So the monotone decreasing interval of Y is x > 3 / 2

It is known that the function f (x) = 1 / 3x ^ 3 + BX ^ 2 + CX (B, C ∈ R) increases monotonically in the interval (- 1,1) and decreases monotonically in the interval (1,3)
1. If B = - 2, find the value of C
2. When B = - 2, find the maximum and minimum value of F (x) in the interval [- 1,3]

When f '(x) = x ^ 2 + 2bx + C = x ^ 2-4x + C (1) x = 1, f' (x) = 0, C = 3 (2) f (x) = x ^ 3 / 3-2x ^ 2 + 3xf '(x) = x ^ 2-4x + 3F' (x) = 0, x = 1, or x = 3, extreme point: F (1) = 1 / 3-2 + 3 = 4 / 3f (3) = 9-18 + 9 = 0, end point value: F (- 1) = - 4 / 3, so maximum value: 4 / 3, minimum: - 4 / 3

Let f (x) = x + | x-a | (1) when a = 2014, find the range of F (x)

When x = f (a) = 2a-a = a
In conclusion, f (x) > = a
That is, the range of F (x) is [a, + ∞)
Substituting a = 2014 into
The range of F (x) is [2014, + ∞)
If the title is correct, the answer is like this, hope to learn to adopt!

The odd function f (x) satisfies f (x + 6) = f (x). When x belongs to (0,3), f (x) = x ^ 2 + 2x, f (2014)=____ ?

2014 = 6 * 335 + 4
So f (2014) = f (6 * 335 + 4) = f (6 * 334 + 4) =... = f (4) = f (- 2)
f(2) = 2*2+2*2 = 8
So f (- 2) = - f (2) = - 8
So f (2014) = - 8

If the odd function f (x) satisfies f (3-x) = - f (x), f (- 1) = - 1, then f (2014)=

It can be seen from the question that f (x) is a periodic function and the period is 3; 2014 = 671 * 3 + 1, so f (2014) = f (1), and f (x) is an odd function,
F (1) = - f (- 1) = 1;

Let f (x) = asin (π x + α) + bcos (π x + β), where a, B, α and β are non-zero real numbers. If f (2013) = - 1, then f (2014)=______ ．

Substituting x = 2013 into: F (2013) = asin (2013 π + α) + bcos (2013 π + β) = - asin α - bcos β = - 1, we get: asin α + bcos β = 1, then when x = 2014, f (2014) = asin (2014 π + α) + bcos (2014 π + β) = asin α + bcos β = 1

Function y = a ^ x-2014 2015 image constant crossing point

The function y = a ^ x is constant over (0,1), so
This question is (0, - 20142014)

The image of LN function
Ln function is what kind of function ah? Why high school textbooks clearly did not learn, do the problem always encounter this kind of function?
How to calculate ln function? Is there a formula
For example: a = log3 is the bottom 2, B = LN2, who is bigger than B and a? How to compare

Ln x is the logarithm of X with E as the base. It is also called the natural logarithm of X. where e = 2.718... Note: ln is used for calculation. You need to use a calculator or look up a table. = = = Baidu Encyclopedia: natural logarithm = = = let a = log (3) (2), B = LN 2, from the bottom changing formula, a = (LG 2)