A few math problems about high school functions, ask the master to solve ,,

Solution:1For any x that belongs to m, there is x+f(x) as an increment function.

y=x is monotonically increasing.

When f(x) is a constant function, it is clear that x+f(x) is an increasing function.

There are two kinds of such f: f(x)=1 and f(x)=2, when f: is the full emission of m n, and there is another f: f(2)=1, f(3)=2, and there are three functions that meet the conditions.

is an odd function defined on r is.

f(x)=-f(-x)

f(x)=-f(4-x)=f(x-4) i.e. f(x+4)=f(x) function, f(x) is a periodic function with a period of 4.

f(2013) = f(4*503+1) = f(1)x belongs to [0,2], f(x)=ax-x

f(1)=a-1

Therefore f(2013) = a-1.

The first question will not, the second question: because he is an odd function, he can also be solved according to the definition f(x)=-f(-x), and the two formulas in the merger problem can be solved, 2013 divided by 4 and the remainder of 1, f(1)=a-1, I don't understand it.

2.Solution: Knowing from the meaning of the question:

f(x)=-f(-x)=-f(4-x), so f(-x)=f(4-x), so f(x)=f(4+x), so the period is 4, so that x=2, then f(2)=f(-2), so f(2)=0, so 2 times a-2 =0, so a=2So f(x)=2x-x, where x belongs to the closed interval of (0,2). So f(2013)=f(4 503+1)=f(1)=1

Solution: (1) It can be seen that y=-x 3 is a monotonically decreasing function on (-infinity, +infinity).

then as long as the conditions are: b=-a3 and a=-b3

The solution is obtained: a=-1 b=1

That is, the interval between the closing function y=-x 3 and the conditional interval is [-1,1](2)The function f(x)=3 4x+1 x=7 4x, (x>0), we can see that the function f(x) is in (0,+infinite) upper monotonically decreasing function, assuming that f(x) satisfies the conditions on [a,b], then there is:

a=7/4b b=7/4a

Get: ab=7 4 Therefore, on (0,+infinity), the condition can be satisfied for any combination that satisfies ab=7 4

Therefore, f(x)=3 4x+1 x(x 0) is a closed function.

First, substituting the p point to the original function equation gives b=2

Then find the derivative at x=1, f'(x)=2x 2-2ax-9, substitute x=1 into -12=2-2a-9

Find a=5 2

x-3 under the root number, the definition domain is x>3, its inverse function is x=y 2+3, and its inverse function definition domain is y>0, that is, the original function value range is (0,+

2.The original function defines the domain as (- 13 4].

y'=2-, the function increments when x<3, 3< x<13 4, the function decreases, so the maximum value is obtained at x=3, and the maximum value y|x=3 = 4, so the range is (- 4]< p>

1. According to the requirements of the defined domain, the root number must not be less than 0, x-1 3>=0, x>=1 3, y [0,+ If f(x)=1 (x-3), x>3, then it is y (0,+ 2, in the same way, 13-4x>=0,x<=13 4,x cannot be greater than 13 4,2x-3 increases with the increase of x, but is limited by the root number, the maximum value of y is: 7 2,y (-7 2).

Solution: (1), since f(x)=f (1)e (x-1)-f(0)x+1 2x, then f (x)=f (1)ex-1-f(0)+x, so that x=1 gets, f(0)=1, then f(x)=f (1)e (x-1)-x+1 2x, f(0)=f (1)e (-1) then f (1)=e, f(x)=e x-x+1 2x, then g(x)=f (x)=e x-1+x, g (x)=e x+1 0, so y=g(x) increases monotonically on x r, then f (x) 0=f (0) x 0, f (x) 0=f (0) x 0, so the monotonically increasing interval of f(x)=e x-x+1 2x is (0, + the monotonically decreasing interval is (- 0).

2), f(x)=e x - x + 1 2 x 2 1 2x 2+ax+b, i.e. e x >=a+1)x +b is true.

a+1)b, we consider (a+1), the case when b has the same sign. You might as well set a+1>0 and b>0

Then e x >=a+1)x +b, let x=1 get a+1+b<=1, so that (a+1)b <=a+1)+b] 2 4=1 4, i.e., the maximum value of (a+1)b=1 4

You can only find the range of A, and A has more than one value ......When a is equal to 0, it holds.

When a is greater than 0, the derivative, and if greater than -1, the value of the time is such that f(x) >0If it is less than -1, then -a+2>0

When a is less than 0, the derivative is still sought, and when 1, 2+a >0When <1, then f(>o do the rest of the calculations yourself, you can't keep asking others.

Derivative, the derivative function f is obtained'(x), order f'(x)=0 to obtain the x of the extreme point, so that the extreme value is greater than 0, the range of a can be solved.