Is SiNx (SiNx + 2) an odd periodic function Is f (x) = SiNx / [SiNx + 2Sin (x / 2)]

Is SiNx (SiNx + 2) an odd periodic function Is f (x) = SiNx / [SiNx + 2Sin (x / 2)]

Function is a pair of available definition validation:
The period should be combined with the period
sinx T=2π
sinx/2 T=4π
T (sum) = 4 π

Is y = | SiNx | odd or even

y=|sinx|=|sin(-x)|
So it's even

Function y = x|x| + Px, X belongs to R, is the function odd or even?

f(x)=x|x|+px
f(-x)=-x|x|-px
f(x)+f(-x)=0
That is, f (x) = - f (- x)
So f (x) is an odd function

Is y = x + 1 odd or even

Basic discrimination method
Y (- x) = y (x) is an even function
Y (- x) = - Y (x) is an odd function
In this question
Y (- x) = - x + 1 non odd and non even

The function y = x | x | x ∈ r satisfies () A. It is an odd function and a minus function B. It is an even function and an increasing function C. It is an odd function and an increasing function D. It is an even function and a minus function

Because the function person = f (x) = x|x|,
ν f (- x) = - x | - x | = - x | x | = - f (x) old man = f (x) is an odd function;
When x ≥ 0, person = f (x) = X2, the symmetry axis of the opening upward is x = 0,
So man = f (x) is an increasing function when x ≥ 0,
Because the monotonicity of the odd function is the same in the symmetric interval about the origin, human = f (x) is an increasing function;
That is, person = f (x) is an odd function and an increasing function
Therefore, C

What else is not only odd but also even except y = 0 and x = 0?

If it is an odd or even function, then f (x) = 0 and the domain is symmetric about the origin
So there are countless
It means that f (x) = 0, and there are a lot of definition fields, as long as they are symmetric about the origin
such as
-1 or X ∈ R
Or X belongs to (- 2, - 1] ∪ [1,2)
wait

Is y = - | x| odd or even Explain why

Even function
f(x)=|x|=|-x|=f(-x)
So it's even

Is y = 1-1|x an odd function or even function

y(-x)=1-1/(-x)=1+1/x
Non odd non even

Is y = √ (x + 1) odd or even

If the domain is defined as x + 1 > = 0, the solution x > = - 1
The definition domain is not symmetric about the origin
It is a non odd and non even function

It is proved that the derivative of derivative even function is odd function and derivative odd function is even function

prove:
Let even derivative function f (x)
Then f (- x) = f (x)
Derivation on both sides:
f'(-x)(-x)'=f'(x)
That is, f '(- x) (- 1) = f' (x)
f'(-x)=-f'(x)
So f '(x) is an odd function
The derivative of an even derivative function is an odd function
Even functions similar to provably differentiable odd functions