The increasing function y = f (x) defined on R has f (X-Y) = f (x) - f (y) for any x, y ∈ R 1 find the value of F (0) and judge the parity of F (x). 2 if f (k times the x power of 3) + F (3 x power - 9 x power - 2) < 0 holds for any x ∈ R, find the range of real number K

The increasing function y = f (x) defined on R has f (X-Y) = f (x) - f (y) for any x, y ∈ R 1 find the value of F (0) and judge the parity of F (x). 2 if f (k times the x power of 3) + F (3 x power - 9 x power - 2) < 0 holds for any x ∈ R, find the range of real number K

f(x-0)=f(x)-f(0)
f(0)=0
f(x+x)=f(x)-f(-x)
f(-x-x)=f(-x)-f(x)
Add the two formulas to get
f(2x)+f(-2x)=0
That is odd function
two
f(k3^x)

It is known that the function y = f (x) defined on R satisfies f (x-1) = - f (x) for any x, when - 1 ≤ X

F (x-1) = - f (x) f (X-2) = - f (x-1) = f (x) | 2 is the period of F (x). When - 1 ≤ X & lt; 1, f (x) = x draws f (x) image. G (x) = f (x) - loga | x | has at least six zeros. F (x) - loga | x | = 0f (x) = loga | x | has at least six intersections. When 0 & lt; a & lt; 1, when loga | x | passes (5, - 1), a = 1 / 5 has six intersections

It is proved by definition that f (x) = x + 1 / X-1 is a decreasing function on (1, + ∞)

Prove: let x1, X2 be on (1, + ∞) and x1f (x2)
X1f (x2) = > F (x) = x + 1 / X-1 is a decreasing function on (1, + ∞)
So the proposition is proved

If f (x) is an odd function and an increasing function on an open interval (negative infinity, 0), and f (2) = 0, then x times f (x)

∵ f (x) is an odd function
It is an increasing function on (- ∞, 0)
It is also an increasing function at (0, + ∞)
∵f(2)=0=-f（-2）
∴f（-2）=0
∵xf(x)＜0
Ψ x ＞ 0 or X ＜ 0
f(x)＜0 f(x)＞0
∵ f (x) is an increasing function on (- ∞, 0)
It is also an increasing function at (0, + ∞)
- 2 ＜ x ＜ 0 or 0 ＜ x ＜ 2
Choose a

Given the function f (x) = A-1 / X (x is greater than 0), we prove that the function y = f (x) is an increasing function from 0 to positive infinity

It is proved that both X1 and X2 are greater than 0, and 00
That is, the function y = f (x) is an increasing function from 0 to positive infinity
The proof is complete

The minimum value of the function f (x) = sinxcosx is ()
A. -1B. -12C. 12D. 1

When x = k π - π 4, K ∈ Z, f (x) min = - 12

What is the minimum value of the function f (x) = sinxcosx?

Let SiNx + cosx = t
And SiNx + cosx = root sign 2Sin (x + α)
｜ t ∈ (- radical 2, radical 2)
There is f (x) = sinxcosx = 1 / 2T ^ 2-1 / 2
So when t = 0, take the minimum
-1/2

For all real numbers x and y, if the function y = f (x) satisfies f (XY) = f (x) f (y), and f (0) ≠ 0, then f (2009) = ()
A. 2008B. 2009C. 1D. 2

Because f (XY) = f (x) f (y), f (0) = f (0) f (0); and ∵ f (0) ≠ 0, ∵ f (0) = 1, ∵ f (0) = f (2009 × 0) = f (2009) f (0) = f (2009) = 1

For all real numbers, if f (XY) = f (x) * f (y), and f (0) is not equal to 0, then f (2003) =?

F (2003 * 1) = f (2003) * f (1), so f (1) = 1, f (0) = f (0) * f (2003) = f (0) * f (1),
f(2003)=f(1)=1

Question 22: finding the minimum positive period of the function y = sin ((π / 3) - 2x) + sin2x requires reasoning and calculation steps The answer is correct
Question 22: finding the minimum positive period of function y = sin ((π / 3) - 2x) + sin 2x
Require reasoning, calculation steps The answer is correct thank you!

Idea: use the following formula to simplify to a sin form
sinA + sinB = 2 sin[(A+B)/2] cos[(A-B)/2]
y＝sin2x＋sin((π/3)－2x)
＝2 sin(π/6) cos(2x-π/6)
＝cos(2x-π/6)
So the minimum positive period is 2 π / 2 = π