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Given the function f (x) = 3x-1 / 3x + 1, find the domain of definition, parity, monotonicity, proof X is the exponent
X is the exponent
Definition field: 3x + 1 is not equal to 0, that is, X ∈ R
Parity: F (x) = 3x-1 / 3x + 1 is changed to f (x) = 1-2 / (3x + 1), so f (- x) = 1-2 / (3-x + 1) = 1-2 × 3x / (3x + 1) (the numerator denominator is multiplied by 3x)
We can get f (x) + F (- x) = 0, so f (x) is an odd function
Monotonicity: F (x) = 1-2 / (3x + 1). Because 3x + 1 ∈ (1, positive infinity), f (x) = 1-2 / (3x + 1) ∈ (- 1,1)
If the domain of F (x) is R and f (x) - 2F (- x) = 3x, ask whether f (x) must be odd or even or not
f(x)-2f(-x)=3x (1)
If you change x to - x, you can get it
Then f (- x) - 2F (x) = - 3x (2)
(1) + 2 * (2) get:
-3f(x)=-3x
f(x)=x
Obviously f (x) is an odd function
The function f (x) defined on R, when x = 3, f (x) = 1, when x is not equal to 3, f (x) = 1 / | x-3 |, if f (x) ^ 2 + AF (x) + B = 0
There are seven different real number solutions. It should be right to find the value range of A,
Let f (x) = t, then T ^ 2 + at + B = 0 has at most two real solutions. The image of F (x) (translate f (x) = 1 / X three units to the right, and then flip the part below the X axis to the top of the X axis) should be symmetric about x = 3, and monotonically decrease on the right side and increase on the left side of x = 3. Therefore, when t > 0 and t ≠ 1, every T will have and only
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