The function f (x) = sin (x + 1) / (1 + x ^ 2), (- ∞)

The function f (x) = sin (x + 1) / (1 + x ^ 2), (- ∞)

f(1)=sin2/2≠0
f(-1)=0
dissatisfaction
F (1) = f (- 1) and f (1) = - f (- 1)
So BC error
also
x^2>=0
x^2+1>=1
0<1/(x^2+1)<=1
-1<=sin(x+1)/(1+x^2)<=1
Bounded
Because 1 / (1 + x ^ 2) is aperiodic
F (x) is also aperiodic
therefore
Option a

Is f (x) = sin (x ^ 2-x) a bounded function, a periodic function, an odd function or an even function

F (x) = sin (x ^ 2-x) is a composite function, | f (x) = sin (x ^ 2-x) ≤ 1, it is a bounded function
It is a periodic function
It is neither odd nor even

Is the function f (x) = sin (x + Pie / 4) - sin (x-pie / 4) odd or even? What is the period Sin is square. The correct answer is odd function, PI

F (x) = sin ^ 2 (x + Pai / 4) - Sin ^ 2 (x-pie / 4)
=(sinxcosπ\4+cosxsinπ\4)^2-(sinxcosπ\4-cosxsinπ\4)^2
=(2sinxcosπ\4)(2cosxsinπ\4)
=Sincos4 Gen
=2 root numbers 2sin2x
It's an odd function
T=π

If the function f (x) = sin ^ 2 (x) - 1 / 2 (x ∈ R), then is f (x) an odd function or even function with minimum positive period

The minimum positive period is π
It's even function

If f (x) is an odd function, G (x) is an even function, and f (x) + G (x) = 1 / X-1, then f (x) =? 1?

f(-x)+g(-x)=1/(-x-1)
∵ f (x) is an odd function and G (x) is an even function
-f(x)+g(x)=1/(-x-1)
2f(x)=1/(x-1)+1/(x+1)
=2x/[(x-1)(x+1)]
f(x)=[x/(x-1)(x+1)]

High school mathematics f (x) is an odd function with period 4. Is f (x + 4) also an odd function?

If f (x) is an odd function with period 4, there will be two conclusions
1 f(-x)= - f(x)
2,f(x+4)=f(x)
Whether f (x + 4) is odd depends on the reasoning process
f(-x+4)=-f(x+4)?
prove:
F (- x + 4) = f (- x) (periodicity) = - f (x) = - f (x + 4), success! The conclusion is true!

F (x) is an odd function with a period of 12. If f (3) = 1, then f (9) =? F (x) is an odd function with a period of 12. If f (3) = 1, then f (9) =? The answer is - 1, but I think it's 1,

This cycle is so long
f(3)=1
f(-9)=1
Odd function: F (- 9) = - f (9)
f(9)=-1

Y = (SiNx + cosx) ^ 2-1, what is the minimum positive period, odd function or even function?

Remove the brackets to get SiNx ^ 2 + cosx ^ 2 + 2sinxcosx-1
=1+2sinxcosx-1
=2sinxcosx
=sin2x
Minimum positive period π
It's an odd function

The function y = SiNx * cosx is a. odd function with minimum positive period π. B. even function with minimum positive period π. C. odd function with minimum positive period of 2 π

y=1/2sin2x
Obvious odd function
T=2π/w=π
So choose a

Y = (SiNx cosx) ^ - 1 is an even function with a minimum period of 2 π, B is an odd function with a minimum period of 2 π, C is an even function with minimum period π, and D is a π odd function

y=1/(sinx-cosx)
=1/[√2sin(x-π/4)]
So t = 2 π
But this is obviously a non odd and non even function