It is known that the odd function f (x) defined on R satisfies that f (x + 2) + F (x) = 0 for any real number x, and if x ∈ [0,1], f (x) = 3x, the value of F (47 / 3) can be obtained

It is known that the odd function f (x) defined on R satisfies that f (x + 2) + F (x) = 0 for any real number x, and if x ∈ [0,1], f (x) = 3x, the value of F (47 / 3) can be obtained

f(x)+f(x+2)=0
Odd function f (x) + F (- x) = 0
So f (x + 2) = - f (x)
-f(x+2)=f(x)
therefore
f(x+4)
=f[(x+2)+2]
=-f(x+2)
=f(x)
f(x+4)=f(x)
So f (47 / 3)
=f(35/3+4)
=f(35/3)
Repeated use
=f(-1/3)
Odd function = - f (1 / 3)
=-3×1/3
=-1

It is proved that on the interval [2,5], the function f (x) = - 2x & sup2; + 3x-1 is decreasing

f（x）＝－2x²＋3x－1
Then f '(x) = - 4x + 3
When x ∈ [2,5], f '(x)

It is proved that f (x) = x square - 2x is a decreasing function in the interval (an infinite, 1)

(1) F (x) is an increasing function in the interval (- ∞, 1]. It is proved that if x 1, x 2 ∈ (- ∞, 1] and x 1 ＜ x 2 are taken, then f (x 1) - f (x 2) = () - () =, ∵, ∵, ∵, and, ∵, ∵, that is, ∵ f (x) is a decreasing function in the interval (- ∞, 1). (2) the image opening of the function is downward, the axis of symmetry is x = 1, ∵ f (x) is in [0,1]

It is proved by deductive reasoning that the premise of the proposition "function f (x) = - x2 + 2x is an increasing function in (- ∞, 1)" is uuuuuuuuuuuuuuuuuu______ ．

Proof: it is necessary to write out the main premise that function f (x) = - x2 + 2x is an increasing function in (- ∞, 1), that is, the proof process of function is an increasing function. Let the domain of definition of function f (x) be I. if for two independent variables X1 and X2 in an interval belonging to the domain of definition, when variable X1 < X2, there is f (x1) < f (x2), then f (x) is in this interval So the answer is: let the domain of definition of function f (x) be I. if for two independent variables X1 and X2 belonging to an interval in the domain, when X1 < X2, there is f (x1) < f (x2), then f (x) is an increasing function in this interval

Given that f (x) = x square + 1 / 2 x, find the monotone interval of F (x) and prove it
X is greater than 0

Using derivative
The derivative of function f (x) = x / (x ^ 2 + 1) is
f'(x)=【1/(x^2+1)-2*x^2】/(x^2+1)^2
When 0

It is proved that f (x) = x / (the square of X + 1) is an increasing function in the interval (- 1,1)
Can we use the definition method to set X1 and X2 instead of derivative

The derivative of F (x) is (1-x ^ 2) / (x ^ 2 + 1) ^ 2 (x ^ 2 + 1) ^ 2 > 0. When (1-x ^ 2) > 0, the increasing solution of F (x) is - 1

F (2x) + F (3x + 1) = 13X ^ 2 + 6x-1

f(x)=ax^2+bx+c
f(2x)+f(3x+1)=a(2x)^2+b(2x)+c+a(3x+1)^2+b(3x+1)+c=13x^2+6x-1
13ax^2+(6a+5b)x+(a+b+2c)=13x^2+6x-1
The coefficients of corresponding terms are equal
13a=13
6a+5b=6
a+b+2c=-1
a=1,b=0,c=-1
f(x)=x^2-1

If the function f (x) = 2ax2-x-1 has exactly one zero point in (0,1), then the value range of a is ()
A. （1，+∞）B. （-∞，-1）C. （-1，1）D. [0，1）

When △ = 0, a = - 18, there is a zero point x = - 2, which is not on (0, 1), so it does not hold. The function f (x) = 2ax2-x-1 has exactly a zero point in (0, 1), that is, f (0) f (1) < 0, that is - 1 × (2a-1) < 0. The solution is a > 1, so a is selected

In the following relations of variables X and Y: ① 3x-2y = 1; ② y = 2x & # 178;; ③ y2 = 3x, where y is the function of X
There are two

A function represents a correspondence between each input value and a unique output value
So, ①, ② yes, ③ No

Given that the function f (x) is an odd function defined on R, when x > 0, f (x) = - 2x & # 178; + 3x + 1, find f (0)

Your problem is probably to find f (x)
When x = 0,
Because the function is odd, so,
f(-0)= -f(0)
f(0)=-f(0)
2f(0)=0
f(0）=0
When X0
Replace every x in F (x) = - 2 & # 178; + 3x + 1 with (- x);
f(-x)= -2(-x)²+3(-x)+1
=-2x²-3x+1
Since f (x) is an odd function, f (- x) = - f (x)
-f(x)=-2x²-3x+1
f(x)=2x²+3x-1
` {2x²+3x-1 (x0)