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Anonymous users2024-02-091)f(x)+f(-x)=0
f(x+3x4)+f(-x+3x4)=0
f(x+12)+f(12-x)=0
f(f(2) let cosx=t -1<=t<=1
sinx2=1-t^2
f(x)=t^2-1-at
4ac-b 2) 4a=-6=-4-a 2 4=-1-a 2 4a = +-2 root 5
b 2a=a 2 does not belong to -1< = t<=1 does not hold.
So when t=1 -a=-6 a=6
t=-1 when a=-6
a=+-6
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Anonymous users2024-02-08f(2) is a bit puzzling, if f(x) here still satisfies the above question: "The periodic function f(x) is an odd function, and one of its periods is 3, f(
, from the odd function we know that f(0)=0
f(x)=-a=0，a=0！！And f(
Obviously, it's not -1 at this point!
Incomprehensible .........
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Anonymous users2024-02-07Because Chai Bi f(x) Lu Bu = f(x+4), -f(x) = f(-x) so f(-3) = f(-3+4) Sui Siju = f(1) = -f(-1) = -1
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Anonymous users2024-02-06f(x) is the odd scattering function of the vertical waiter.
So f(1)=-f(-1)=-3
f(x) is a periodic function with period 4.
The fiber is infiltrated to. f(5)=f(1)=-3
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Anonymous users2024-02-05The problem is wrong, it should be f(1)=-1, otherwise it can't be solved.
Because it's an odd function.
The finch is f(-1)=-f(1)=1
And because of the minimum positive period.
is 3, and 101 to -1 is exactly (101+1) 3=34 cycles.
F(101)=1
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Anonymous users2024-02-04The odd ash scatter function f(x) is a period of 4 and the period of the trembling period.
f(-2)=-f(2), i.e. f(-2+4)=-f(2), then f(2)=-f(2), i.e. f(2)=0, so the answer is: 0
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Anonymous users2024-02-03Because f(x) is a periodic function with 5 as the period of the rubber disturbance, and f(-1)=1, f(-1+5)=f(4)=1
f(x) is the Lie chain of the beam-empty odd function.
So f(-x) = f(x), f(-4) = f(4) = -1
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Anonymous users2024-02-02It is known that the functions f(x) and x r are odd functions with a period of 4 in the cavity, and f(3) 1f(-1)=f(-1+4)=f(3)=1
The number of odd circle digs f(-1)=-f(1)=1
So Kai laughs f(1)=-1
f（2013）
f(2012+1)f(1)
I just answered it for someone else yesterday, and I copied it directly and changed it slightly, but you didn't have a third question. If you look at it in general, the approach is the same, very similar, but in fact, a question has been slightly changed. If you are interested, you can click on the third question I answered to take a look. >>>More
1) f(x)=-x is subtracted on r, so the condition is satisfied, and when x [-1,1], the set of values of f(x) is also [-1,1], and the condition is satisfied. >>>More
p [3 4,+ f(x) is an even function, and on [0,+ is a subtraction function. >>>More
Let f(x)=ax squared + bx+c,,, because f(0)=1, substituting 0 into c=0, that is, f(x)=ax squared + bx, and because f(x+1)=f(x)+x+1, substituting f(x)=ax squared + bx into this equation, we get ax squared + (b+1)x+1=ax squared + (2a+b)x+a+b, and the solution of a=b=b+1 from the principle of constant eqation gives a=b=one-half. Therefore, f(x) = one-half x square minus one-half x I believe that the landlord has done the second question, and I wish the landlord learning progress.
The idea is for you, knowing that the coefficient of the quadratic term is a, then let f(x)=ax +bx+c >>>More