Given the function f (x) = a-22x + 1 (1), if the function is odd, find a; (2) judge the monotonicity of F (x) on R, and prove your conclusion

Given the function f (x) = a-22x + 1 (1), if the function is odd, find a; (2) judge the monotonicity of F (x) on R, and prove your conclusion

(1) ∵ function is odd, ∵ f (0) = 0, ∵ A-1 = 0, ∵ a = 1, ∵ A is 1. (2) according to (1), f (x) = 1-22x + 1, ∵ the function is an increasing function on R, and it is proved as follows: let X1, X2 ∈ R, and X1 < X2, f (x1) - f (x2) = 1-22x1 + 1 − 1 + 22x2 + 1, = 2 (2x1 − 2x2) (2x1 + 1) (2x2 + 1), ∵ X1 < X2, ∵ 2x1 − 2x2 < 0, ∵ f (x1) - f (x2) < 0, M the function is an increasing function on R The function is an increasing function on R

Judge the monotonicity of function FX = (x ^ 2-2x + 5) / X-1 on (3, +) and prove it

Judge the monotonicity of function f (x) = (X & # 178; - 2x + 5) / (x-1) on (3, + ∞) and prove it
Since f '(x) = [(x-1) (2x-2) - (X & # 178; - 2x + 5)] / (x-1) & # 178; = (X & # 178; - 2x-3) / (x-1) & # 178; = (x-3) (x + 1) / (x-1) & # 178; > 0 is constant on (3, + ∞), f (x) increases monotonically on (3, + ∞)
This reasoning process is very rigorous, and there is no need to prove it separately

Let y = f (x) (x belongs to R) satisfy f (x1) + F (x2) = f (x1 * x2) for any real number x1, X2, and prove that f (x) is an even function

For any x, f (x) + F (0) = f (x * 0) = f (0), so f (x) = f (0) = constant function, that is, for any x, f (- x) = f (x) = constant function is satisfied

Given the even function f (x), for any x 1, x 2 belongs to R, f (x 1 + x 2) = f (x 1) + F (x 2) + 2 (x 1x 2) + 1, find the value of F (2)

f(4)=f(2)+f(2)+f(2*2)+1
=2f(2)+f(4)+1
2f(2)=-1
f(2)=-1/2

Symmetry line of trigonometric function
The graph of function f (x) = 4sin (2x + 3 / 2 π)
A. On X-axis symmetry B. on origin symmetry C. on Y-axis symmetry D. on line x = π / 2 symmetry
I think we should choose D
What do you say to write? It's totally confusing.

The axis of symmetry of sin is where sin reaches its maximum
That is sin (2x + 3 / 2 π) = ± 1
therefore
2x+3/2π=kπ+π/2
x=kπ/2-π/2
So k = 2, x = π / 2
So D is right

Symmetry of trigonometric function
F (x) = sin2x + acos2x is symmetric with respect to the line x = - π / 8,
So f (x) = f (2 (- π / 8) - x) = f (- π / 4-x)
Why is this? What conclusion is this? F (x) = f ((2 times the axis of symmetry) - x) I don't quite understand

Let t = A-X, then x = A-T, f (A-T) = f (a + T) is obtained from F (x) = f (2a-x), that is, f (A-X) = f (a + x) only means that the function values of the values which are far from the left and right of a are equal, so f (x) is symmetric with respect to x = a

Trigonometric function symmetry
Why "for sine function y = asin (ω x + Φ), let ω x + Φ = k π + π / 2 solve x to get the axis of symmetry, let ω x + Φ = k π solve x is the abscissa of the center of symmetry, and the ordinate is 0"? I want to know the reason why ω x + Φ = k π + π / 2. Thank you very much!

From the image of y = SiNx, we can see that the axis of symmetry is the set of maximum and minimum values of Y, so x = k π + π / 2
And the center of symmetry is the set of intersection points of y = SiNx and X axis

Judging the parity of function f (x) = cos (5 / 2 π + 2x)

cos（5/2π+2x）=cos(2π+1/2π+2x)=cos1(/2π+2x)=-sin2x
It is an odd function

Judging the symmetry of F (x) = 2x ^ 4 + x ^ 2 by using the parity of function

F (x) is an even function, so f (x) is symmetric about the Y axis

Given the function f (x) = 2x-a / x, and f (1) = 3 (1), find the value of a (2) judge the parity of the function (3) find the maximum value of F (x) on [2,4]

(1)f(x)=2x-a/x
x=1 2-a=3
a=-1
(2)f(x)=2x+1/x
f(-x)=-2x-1/x=-(2x+1/x)=-f(x)
Odd decreasing function
(3) Function in (0, + ∞)
f(x)=x+1/x>=2
Then x > = 1 is an increasing function
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