F (x) is an odd function defined on R, and f (x + 2) = - f (x). When 0 ≤ x ≤ 1, f (x) = x ^ 2 + X ① find the period ② find the expression at - 1 ≤ x ≤ 0 ③ Find f (6.5)

F (x) is an odd function defined on R, and f (x + 2) = - f (x). When 0 ≤ x ≤ 1, f (x) = x ^ 2 + X ① find the period ② find the expression at - 1 ≤ x ≤ 0 ③ Find f (6.5)

F (x + 4) = f (x + 2 + 2) = - f (x + 2) = f (x) ① find the period as 4
② If - 1 ≤ x ≤ 0, then 0 ≤ - x ≤ 1 f (x) = - f (- x) = - (x ^ 2-x) = - x ^ 2 + X
② Find the expression f (x) = - x ^ 2 + X at - 1 ≤ x ≤ 0

If the function f (x) defined on R has countless real numbers x and satisfies f (x + 2) = f (x), is f (x) a periodic function

The answer upstairs has a question
For the function f (x) defined on R, there are countless real numbers x that satisfy f (x + 2) = f (x). It can not be concluded that f (x) is a periodic function with 2 as the cycle! Because countless real numbers x may not fill the whole real number field. If it is changed to: "the function f (x) defined on R satisfies f (x + 2) = f (x) for any real number x", then f (x) is a periodic function with period 2

F (x) is an odd function with period of 3 defined on R, and f (2) = 0, then how many real solutions do the equation f (x) = 0 have in the interval (0,6)

Four. I haven't counted it for a long time
f(2)=0.
And f (x) = f (- x)
F (2 + 3K) = 0, K is any integer
Moreover, | 2 + 3K | is in the interval (0,6)
So, k = - 2, - 1,0,1
That is, there are four solutions

F (x) is an odd function with period of 3 defined on R, f (2) = 0, then the number of solutions of equation f (x) = 0 in interval (0, 6) () A. It's three B. It's four C. It's five D. More than 5

∵ f (x) is an odd function with a period of 3 defined on R, f (2) = 0. If x ∈ (0, 6), then f (5) = f (2) = 0. According to f (x) is an odd function, then f (- 2) = f (2) = 0, then f (4) = f (1) = f (- 2) = 0. Also, the function f (x) is an odd function defined on R

F (x) is an even function defined on R with period 3, and f (2) = 0, then the number of solutions of equation f (x) = 0 in interval (0,6) is at least several

F (2) = 0 because it is an even function f (- 2) = 0
So f (2 + 3) = f (5) = 0
f(-2+3)=f(1)=0
f(1+3)=f(4)=0
So four

It is known that the domain of function f (x) is all real numbers of X ≠ 0. For any X1 and X2 in the domain, f (x1 * x2) = f (x1) + F (x2) And when x > 1, f (x) > 0, f (2) = 1 Verification: F (x) = f (- x) F (x) is an increasing function on (0, + ∞) Solving inequality f (|x| + 1) < 2

(1) Because x is not equal to 0, if f (x1 * x2) = f (x1) + F (x2), then X1 = x2 = 1, then f (1 * 1) = f (1) + F (1), then f (1) = 0; if X1 = x2 = - 1, then f (1) = f (- 1) + F (- 1), that is, 2f (- 1) = 0, then f (- 1) = 0; if x2 = - 1, X1 = x, then f (- x) = f (x) + F (- 1), then f (- x) = f (x)