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If f (x) satisfies 2F (x) + F (- x) = 3x + 2, then f (2)=________ ? Can you carry two numbers at will, for example: 2f（2）+f（-2）=8 2f（-2）+f（2）=-4 Then, the equations are solved to obtain f (2) = 4
∵ 2F (x) + F (- x) = 3x + 2 ᙽ change x into - x to get 2F (- x) + F (x) = - 3x + 2
Let f (x) be odd function, G (x) be even function, and f (x) + G (x) = 2x ^ 2-3x + 1, find the analytic formula of y = f (x), y = g (x)
Solve a problem to analyze the conditions, this kind of problem is a routine. Think clearly one, all types will be
Condition 1: F (x) is odd function and G (x) is even function
Then f (- x) = - f (x); G (- x) = g (x) (1)
[image understanding of definition; even function eats up negative sign, and odd function spits out]
Condition 2: F (x) + G (x) = 2x ^ 2-3x + 1 (2)
Replace X in (2) with - X
f（-x)+g（-x)＝2x^2+3x+1 （3）
Because (1) f (- x) = - f (x); G (- x) = g (x), (3) becomes
- f（x)+g（x)＝2x^2+3x+1 （4）
(2) The sum of (4) is: G (x) = 2x ^ 2 + 1
(2) By subtracting (4), f (x) = - 3x
The solution is over
If f (x) is an odd function, it only contains the odd term of X, that is - 3x
If G (x) is an even function, it will only contain the even degree term of X, that is, 2x ^ 2 + 1 [1 is regarded as the zero degree of x]
If even function f (x) = x ^ 3-8 (x > = 0), then {x | f (X-2) > 0}= A{X|X4} B{x|x4} C{X|X6} D{X|X2} I think it's strange, but its condition is x > = 0? Should the other side be axisymmetric?
This is a piecewise function
When x > = 0, the analytic formula is given, we can find x = 0, the solution (X-2) ^ 3-8 > 0, and the union of x > = 2 (x > 4)
At x = 2, Union (x) is obtained
Given that even function f (x) satisfies f (x) = x3-8 (x ≥ 0), then the solution set of F (X-2) > 0 is______ .
Because f (x) is an even function,
When x ≥ 0, f (x) = x3-8 is an increasing function,
When x ≤ 0, f (x) is a minus function;
∵f（x-2）＞0=f（2），
So we can get: | X-2 | 2,
The solution is: x < 0, or x > 4
So the answer is: (- ∞, 0) ∪ (4, + ∞)
If f (x + 8) is an even function, then f (x + 8) = f (- X-8) or F (- x + 8) It's easy to confuse,
F (x + 8) is to move f (x) to the left by 8 units
F (x + 8) symmetric about y-axis
Obviously, moving f (x + 8) eight units to the right is f (x)
Then the axis of symmetry has to move eight units to the right
So the axis of symmetry f (x) is x = 8
So there is f (8 + x) = f (8-x)
So it should be f (- x + 8) right
If f (x + 8) is an even function, why is there f (x + 8) = f (- x + 8) Online all say definition comes - I preview, really not easy to learn Can people help, by the way, what is the meaning of 8 in brackets They say that 8 is the axis of symmetry. If it's about the Y axis, it's not f (x + 0) = f (- x + 0) The definition of F (x) = f (- x) is about y axis symmetry and omit 0? Sweat, so confused,
If f (x + 8) is even, why f (x + 8) = f (- x + 8)? Because f (x) itself is a periodic function, then: F (x) = f (x + 16), period 16; f (x + 8) = f (X-8) = f {- (X-8)} = f (- x + 8)
It is known that f (x + 1) is even function and f (x) = x ^ 2 + X when x is less than or equal to 1
Sorry, I made a mistake just now
∵ f (x + 1) is an even function
∴f(x+1)=f(-x+1)
∴f[(x-1)+1]=f[-(x-1)+1]
∴f(x)=f(2-x)
∵ when x ≤ 1
f(x)=x^2+x
When x > 1
2-x
When f (x) is known to be even and X ≤ 0, f (x) = 1 + X 1 − x, find (1) The value of F (5); (2) The value of X when f (x) = 0; (3) The analytic formula of F (x) when x > 0
（1）f（5）=f（-5）=1−5
1+5=-4
6=-2
Three
(2) When x ≤ 0, f (x) = 0 is 1 + X
1−x=0，
ν x = - 1, and f (1) = f (- 1),
When f (x) = 0, x = ± 1
(3) When x > 0, f (x) = f (- x) = 1 − X
1+x，
When x > 0, f (x) = 1 − X
1+x．
It is known that the function f (x) whose domain of definition is R is even function. When x is greater than or equal to 0, f (x) = x + 1 / X?
Let x0, so there is f (- x) = (- x) / (- x + 1)
If f (- x) = f (x), then x = 0)
f(x)=-x/(-x+1),(x
It is known that the function f (x) is an odd function defined on R. when x ≥ 0, f (x) = x (1 + x), the graph of function f (x) is drawn, and the analytic formula of function f (x) is obtained
∵ when x ≥ 0, f (x) = x (1 + x) = (x + 1)
2）2-1
4，
F (x) is an odd function defined on R,
When x < 0, - x > 0,
f（-x）=-x（1-x）=（x-1
2）2-1
4=-f（x），
∴f（x）=-（x-1
2）2+1
Four
∴f（x）=
(x+1
2) 2-1
4 x≥0
-(x-1
2) 2+1
4 x＜0
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