Given that the function f (x) = 2Sin ^ 2 x + sin2x, X belongs to [0, π], find the set of X whose f (x) is positive When I look up this problem, I found that there is a problem whose value range is not the same, so steps should be taken, not copied

Given that the function f (x) = 2Sin ^ 2 x + sin2x, X belongs to [0, π], find the set of X whose f (x) is positive When I look up this problem, I found that there is a problem whose value range is not the same, so steps should be taken, not copied

If f (x) = 2Sin ^ 2 x + sin2x = 2Sin ^ 2 X-1 + sin2x + 1 = - cos2x + sin2x + 1 = sin2x-cos2x + 1 = sin2x-cos2x + 1 = sin2x-cos2x + 1 = sin2x (2x - π / 4) + 1 because x ∈ [0, π], so 2x - π / 4 ∈ [- π / 4,7 π / 4] in order to make f (x) > 0, we must make sin (2x - π / 4) > - (√ 2) / 22x - π / 4 ∈ [- π / 4,4 ∈ [- π / 4,4,4,π / 4 ∈ [- π / 4,7 π / 4,4,4,5 π / 4] x ∈ [0,3 π /

Among the following functions, what is the even function with the smallest positive period? Ay=sin2x By=cos2x Cy=cosx Dy=tanx

It's by = cos2x
The first cycle: the period of sinax and cosax is 2
The Tanax period is U / A
Exclude CY = cosx
Even function f (x) = f (- x)
Exclude ay = sin2x and Dy = TaNx

If the function f (x) = the square of AX + BX + 3A + B is even function, the definition domain is [a-1,2a], find the value range of F (x)

F (x) = ax 2 + BX + 3A + B is even function
Then the definition domain is symmetric about the origin, that is, A-1 = - 2A, and a = 1 / 3 is obtained
f(-x)=ax²-bx+3a+b
f(x)=f(-x)
So B = - B; b = 0
f(x)=x²/3+1 x∈[-2/3,2/3]
Function range [1,31 / 27]

Given that even function f (x) is an increasing function on (0, + ∞), is f (x) an increasing function or a decreasing function on (- ∞, 0)? Please prove your conclusion

F (x) is a decreasing function on (- ∞, 0). The reasons are as follows: let X1 < x2 < 0, then - X1 > - x2 > 0 (2) because f (x) is an increasing function on (0, + ∞), so f (- x1) > F (- x2) (4 points) and f (x) is even function, so f (x1) > F (x2) （6...

It is known that f (x) = {(6-A) x-4a (x)

Give your kids
But I'm most likely to dream about someone who's not there,
Time, seize it and capture it into my dark room——
You make them float and sink in gold,
Know this kind of reciprocation
Where are you comfortable? Ha ha

Let f (x) and G (x) be defined in R and X is not equal to plus or minus 1, f (x) is even function, G (x) is odd function, and f (x) plus g (x) = 1 / (x-1), what is the analytic formula of f (x) and G (x) F (- x) = f (x), G (- x) = - G (x), substituting f (- x) - G (x) = 1 / (- x-1), The key is why f (- x) - G (x) = 1 / (- x-1),

In the equation f (x) + G (x) = 1 / (x-1), if every x is changed into - x, then there is
f(-x)+g(-x)=1/[(-x)-1]
And [- x] and [- x] = (- 1 /) (- 1 /)
Therefore, f (x) - G (x) = 1 / (- x-1) = - 1 / (x + 1)
f(x)+g(x)=1/(x-1) (1)
f(x)-g(x)=-1/(x+1) (2)
(1) The results show that 2F (x) = 1 / (x-1) - 1 / (x + 1) = 2 / (x ^ 2-1), f (x) = 1 / (x ^ 2-1)
(1) (2) 2 g (x) = 1 / (x-1) + 1 / (x + 1) = 2x / (x ^ 2-1), G (x) = x / (x ^ 2-1)

Let f (x) be an odd function defined on R, and f (x+2) =-f (x). When -1 ≤ x ≤ 1, f (x) =x is a three power function It is proved that (1) the line x = 1 is a symmetric axis of the image of function f (x) (2) When x belongs to [1,5], find the analytic formula of F (x)

(1) Because it is an odd function, f (x + 2) = - f (x) = f (- x)
If x is replaced by X-1, f (1 + x) = f (1-x)
So the line x = 1 is an axis of symmetry of the image of function f (x)
(2) Because f (x + 2) = - f (x) = f (X-2)
So 4 is a period, because when - 1 ≤ x ≤ 1, f (x) = x ^ 3
So when 3 ≤ x ≤ 5, f (x) = x ^ 3
Because when 1 ≤ x ≤ 3, f (x + 2) = - f (x) also holds, and 3 ≤ x + 2 ≤ 5
So f (x + 2) = (x + 2) ^ 3 = - f (x) / / 1 ≤ x ≤ 3
So when 1 ≤ x ≤ 3, f (x) = - (x + 2) ^ 3
To sum up, when 3 ≤ x ≤ 5, f (x) = x ^ 3; when 1 ≤ x ≤ 3, f (x) = - (x + 2) ^ 3
I wonder if you are satisfied with my answer

If the function f (x) is π 2 is a periodic function, and f (π) 3) If = 1, then f (17 6π）=（　　） A. 1 B. 2 C. 3 D. 4

∵ the function f (x) is defined by π
2 is a periodic function, and f (π)
3）=1，
∴f（17
6π）=f（4×π
2+5π
6）=f（5π
6）=f（5π
6-π
2）=f（π
3）=1．
Therefore, a

It is known that the odd function f (x) and even function g (x) whose definition domain is r satisfy the x power of F (x) + G (x) = 10, and find the analytic expressions of the functions f (x) and G (x)

f（x）+g（x）=10^x
F (- x) + G (- x) = 10 ^ - x, that is - f (x) + G (x) = 10 ^ - X
Are simultaneous equations meeting the ball

Even function f (x) defined on R is monotonically increasing at (- -, 0) if f (2a ^ + A + 1)

From the invariance of 2A2 + A + 1 > 0, 3a2-2a + 1 > 0, we can know that 2A2 + A + 1 > 3a2-2a + 1, that is, 0 < a < 3 (∵ f (x) is an even function, ᙽ on (0, + ∞) is a minus function)