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If the even function f (x) over the definition field r has f (x) = - f (x + 3 / 2) for any real number x, and f (- 1) = 1, f (0) = - 2, then f (1) + F (2) +... " The value of F (2012) is
F (x) is the even function f (x) is the even function f (- x) = - f (x) f (1) = - f (- 1) = - 1 ∵ f (x) = - f (x + 3 / 2), f (x + 3 / 2) = f (x + 3 / 2) = f (x + 3 / 2) = f (x + 3 / 2) + 3 / 2] = - f (x + 3 / 2) + 3 / 2] = - f (x + 3 / 2) = f (x + 3 / 2) = f (x), f (x) is a periodic function, and the period is 3 ȿ f (2) = f (- 1) = 1F (3) = f (3) = f (0) = 2F (4) = f (1) = f (1- 1F (5) = f (2) = 1. F (1) + F (2) + F (
If the even function f (x) over the definition field r has f (x) = - f (x + 3 / 2) for any real number x, and f (- 1) = 1, f (0) = - 2, then f (1) + F (2) +... " The value of F (2011) is
F (3) = f (3 / 2 + 3 / 2) = - f (3 / 2) = f (0) ---- f (3K) is the same
F (4) = f (3 / 2 + 3 / 2 + 1) = - f (3 / 2 + 1) = f (1) ---- f (3K + 1)
F (5) = f (3 / 2 + 3 / 2 + 3 / 2 + 3 / 2-1) = - f (- 1) ---- f (3K + 2) is the same
…… Word limit, depressed
Questioning
If f (x) is an even function defined on R and is an increasing function when x is greater than or equal to 0, then f (x)
Because f (- x) = f (x),
So f (x)
F (x) = | 1 / 2 of | A-X | (a ∈ R)... Determine the value of real number a, so that the function f (x) is an even function in its domain of definition? ..
Let's assume that the function f (x) is the even function in its definition domain, then f (- x) = f (x-1) | (AX-1) | / | | f (- x-x) = a + 1 + 1 / X | (AX + 1) | (AX + 1) / x | (x + 1) | (AX + 1) | (AX + 1) | (AX + 1) | (AX + 1) | (AX + 1) | (AX + 1) | (AX + 1) | (AX + 1) | / | (AX + 1) | / | (AX + 1) | / | / | (AX + 1) | (AX when we remove the absolute pair, ax + 1 = AX-1 is obviously not true
If the definition domain of a function is (4a-3,3-2a) and y = f (2x-3) even function, then the value of real number is Find the value of real number a! To correct one point, the domain is 4a-3 3-2a ^ 2
From 4a-3 < x < 3-A,
When 4a-3 < 2x-3 < 3-A
4a＜2x＜6-a²
When 2A ＜ x ＜ 3-A 2 / 2, it is even function,
From 3-A 2 / 2 > 2a,
Results: a 2 + 4a-6 < 0
（a-+2+√10）（a+2-√10）＜0
-2-√10＜a＜-2+√10.
If the definition domain of even function is interval (- 4A, A2 + 3), then real number a=
The image of even function is symmetric about y-axis, and the definition domain is also symmetric about the origin, so the addition of two points in the interval is equal to zero, - 4A + A2 + 3 = 0, and a = 1 or 3 is obtained, which is substituted into the detection interval
Let f (x) be an even function defined on the real number set R, and be a decreasing function on (- ∞, 0), and f (2A2 + A + 1) > F (3a2-2a + 1), and find the value range of A
∵ f (x) is an even function on R, and it is a decreasing function on (- ∞, 0), and ∵ f (x) is an increasing function on (0, + ∞). It is also ? 2A2 + A + 1 = 2 (a + 14) 2 + 78 > 0, 3a2-2a + 1 = 3 (A-13) 2 + 23 > 0, ? inequality f (2A2 + A + 1) > F (3a2-2a + 1), equivalent to 2A2 + A + 1 > 3a2-2a + 1
Let f (x) be an even function defined on the real number set R, and be a decreasing function on (- ∞, 0), and f (2A2 + A + 1) > F (3a2-2a + 1), and find the value range of A
∵ f (x) is an even function on R, and it is a decreasing function on (- ∞, 0), and ∵ f (x) is an increasing function on (0, + ∞). It is also ? 2A2 + A + 1 = 2 (a + 14) 2 + 78 > 0, 3a2-2a + 1 = 3 (A-13) 2 + 23 > 0, ? inequality f (2A2 + A + 1) > F (3a2-2a + 1), equivalent to 2A2 + A + 1 > 3a2-2a + 1
Given that f (x) = AX2 + BX + 3A + B is an even function and its definition domain is [A-1, 2A], then the locus of point (a, b) is () A. Point B. Straight line C. Line segment D. Ray
∵ the domain should be symmetric about the origin,
So there is A-1 = - 2A,
A = 1
3．
And ∵ f (- x) = f (x) is constant,
That is, AX2 + BX + 3A + B = AX2 BX + 3A + B
∴b=0．
The point (a, b) is (1)
3，0）
Therefore, a
It is known that the function f (x) = AX2 + BX + 3A + B is even function, and its definition domain is [a-1,2a]. (1) find the value of a and B; (2) find the maximum value of function f (x) in its definition domain
The function f (x) = AX2 + BX + 3A + B is even function
Then f (x) = f (- x)
So B = 0
If the definition domain of even function is symmetric about 0, then A-1 = - 2A
A = 1 / 3 is obtained
So f (x) = AX2 + BX + 3A + B = 1 / 3x ^ 2 + 1
Its definition domain is [- 2 / 3,2 / 3]
Obviously, when x = - 2 / 3 or 2 / 3, the maximum value is 31 / 27
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