I have two questions. 1. Do odd functions have f (0) = 0? Why? two Given function f (x) has f (y + x) = f (x) + 2Y (x + y) for any real number x, y Cut f (1) = 1, find the analytic expression of F (x) That's what I did F（1+Y）=F（1）+2Y（1+Y） Let 1 + y = T. replace y = T-1 Replace the above formula to get 2T square - 2T = 1 But the answer is 2x squared - 1 What's wrong with me, I know algebra, but why not?

I have two questions. 1. Do odd functions have f (0) = 0? Why? two Given function f (x) has f (y + x) = f (x) + 2Y (x + y) for any real number x, y Cut f (1) = 1, find the analytic expression of F (x) That's what I did F（1+Y）=F（1）+2Y（1+Y） Let 1 + y = T. replace y = T-1 Replace the above formula to get 2T square - 2T = 1 But the answer is 2x squared - 1 What's wrong with me, I know algebra, but why not?

1.Y
2. I think you can't: F (1 + y) = f (1) + 2Y (1 + y),
You try algebra
Pay attention to the basics!

It is known that the domain of function f (x) = LG (x * 2 + ax + b) is set a, and the domain of function g (x) = √ KX * 2 + 4x + K + 3 is set B. if (CRA) ∩ B = B, (CRA) ∪ B = [- 2,3], the value of real number a, B and the value range of real number k are obtained
Let f (x) be a function defined on R. for X ∈ R, f (x) + F (x + 2) = 0. When - 1 ≤ x ≤ 1, f (x) = the third power of X (1). For 1 ≤ x ≤ 5, the analytic expression of function f (x) (2) if a = {x │ f (x) > A, X ∈ r}, and a ≠ empty set, the value range of real number a is obtained

Senior one math function exercise-1
1) The function f (x) = x square + 1 is a decreasing function on (- ∞, 0);
2) The function f (x) = 1-1 / X is an increasing function on (- ∞, 0)

1. Because the derivative of the function f (x) is 2x,
Because x is on (- ∞, 0)
So the derivative of function f (x) is less than 0
So the function f (x) = x squared + 1 is a decreasing function on (- ∞, 0)
2. Because the derivative of the function f (x) = 1-x is 1 / (1-x) ^ 2, which is greater than 0, X is not equal to 1
So the function f (x) = 1-1 / X is an increasing function on (- ∞, 0)
Method 2: it's OK to use the method of definition

1. Evaluation: 2 √ 3 × cubic root 1.5 × sixth root 12
2. Calculation: ① (A & sup2; - 2 + A ^ - 2) / (A & sup2; - A ^ - 2)
3. If a + A ^ - 1 = 3, find the values of a ^ (1 / 2) - A ^ (- 1 / 2) and a ^ (3 / 2) - A ^ (- 3 / 2)

1、
Original formula = 2 * 3 ^ (1 / 2) × (3 / 2) ^ (1 / 3) × (2 & sup2; × 3) ^ (1 / 6)
=2*3^(1/2)×3^(1/3)/2^(1/3)×2^(1/3)×3^(1/6)
=2^(1-1/3+1/3)×3^(1/2+1/3+1/6)
=2×3
=6
2、
The original formula = [A-A ^ (- 1)] & sup2; △ {[a + A ^ (- 1)] [A-A ^ (- 1)]
=[a-a^(-1)]/[a+a^(-1)]
Up and down a
=(a²-1)/(a²+1)
3、
[a^(1/2)-a^(-1/2)]
=a-2+a^(-1)
=3-2
=1
So a ^ (1 / 2) - A ^ (- 1 / 2) = ± 1
a^(3/2)-a^(-3/2)
=[a^(1/2)-a^(-1/2)][a+1+a^(-1)]
=±1×(3+1)
=±4