Two high level function problems, two high level function math problems

1.The function f(x) = (m-1) x + mx + 3 (x r) is an even function.

m=0f(x)=-x²+3 (x∈r)

The monotonic reduction interval for f(x) is (0,+

2.The monotonic increase interval of the function f(x)=4x x +1 is (-1,1)(m,2m+1) which is a subset of (-1,1).

m>=-1 and 2m+1<=1

1<=m<=0

1. If it is an even number, then m=0, that is, f(x)= x 3, and the subtraction interval is (0];

2. f(x)=4x(x 1), then f'(x)=[4(x 1) 8x ] [x 1] =[4 4x] [x 1], the increasing interval is [ 1,1], then: 1 m<2m 1 1, the solution is: 1

1.The even function is that the axis of symmetry is 0, m=0, f(x)=-x 2+3, the opening is downward, and the monotonic reduction interval is x>=0;

f(x)=4/g(x)

m>-1, when 2m+1<0, g(x)<0 and subtract the function, then f(x) increases the function to obtain m<-1 2;

When m>0, g(x)>0, and subtract the function on (0,1), and 2m+1-m=m+1>=1, so m=0

m (-1, -1 2) or m = 0

1.The even function is that the axis of symmetry is 0, m=0, f(x)=-x 2+3, the opening is downward, and the monotonic reduction interval is x>=0;

2. The monotonic increase interval of the function f(x) = 4x x +1 is (-1,1), and only (m,2m+1) is a subset of (-1,1).

m>=-1 and 2m+1<=1, i.e., -1<=m<=0

1. Translate one unit to the right.

2. (1) Because f(xy)=f(x)+f(y), let x=1 2, y=1

Therefore f(1 2) = f(1 2) + f(1).

So f(1)=0

2)f(1)=f(2)+f(1/2),f(2)=-1

f(4)=f(2)+f(2)=-2

For 0 less than x and less than y, both f(x) are greater than f(y), which means that f(x) is a subtractive function f(-x) + f(3-x) -2

f((-x)*(3-x)) 2

f(x^2-3x) ≥2

f(x^2-3x) ≥f(4)

i.e. x 2-3x 4

x^2-3x-4≤0

x-4)(x+1)≤0

1 x 4 Finally, note that the two defining domains are (-x)>0 and (3-x)>0, and the final answer is -1 x<0

1.Shift left units 2(1) Let x=y=1, then f(1)=0 (2) Let x=2, y=then f(1)=f(2)+f(then f(2)=-1;Let x=y=2 then f(4)=2f(2) =-2 can be huahua f(x squared -3x) -2=f(4) and for 0 less than x less than y, both f(x) is greater than f(y).

Then x squared -3x 4 gives -1 x 4Finish.

f(x)=x^2+4x+3

f(ax+b)=(ax+b)^2+4(ax+b)+3=a^2x^2+(2ab+4a)x+ b^2+4b+3

x^2+ 10 x+ 24

Pending coefficients: a 2=1 (2ab+4a)=10 b 2+4b+3=24SO: a=1 b=3 or a=-1 b=-7SO: 5a-b=2

2.In the first case: f(x) is not a quadratic function i.e. (5-a)=0 i.e. a=5

f(x)=-6x +10 When x belongs to r, it is impossible to be constant (round) In the second case, f(x) is a quadratic function, i.e. a= 5, when 5-a>0 a<5, the opening is upward, as long as deierta<0 is enough, i.e. 36-4(5+a)(5-a)<0 is connected: at 45, the opening is downward, and there must be a negative value (round).

In sum -4

1. Because f(x)=x(squared)+4x+3, f(ax+b)=(ax+b)(squared)+4(ax+b)+3

And f(ax+b)=x(square)+10x+24, so a=1 or -1, so f(ax+b)=(x+b)(squared)+4(x+b)+3=x(squared)+2(b+2)x+b(squared)+4b+3 or f(ax+b)=(x+b)(squared)+4(-x+b)+3=x(squared)+2(-b+2)x+b(squared)+4b+3, so when a=1, 10=2(b+2) or when a=-1, 10=2(b-2), i.e. when a=1, b=3; When a=-1, b=7

1、（1）

f(m)=m*2 (am-2)=m, which is arranged to form a=(m+2) m, and a is a positive integer, so the value of a is 2 or 3.

1) When a=2, m=2:

f(-m)-(1/m)=-1/6<0

2) When a=3, m=1:

f(-m)-(1 m)=-3 4>0, rounded off so a=2, m=3

f（x）=x*2/(2x-2)

2）a1=1………Calculate the formula for finding the pattern and writing the general term.

2. I think this question is a bit problematic (personal opinion, not necessarily accurate), according to the title, we should write f(-x+5)=f(x-3), find the values of a and b by the contrast coefficient method, and find the c value by (b-1)*2-4ac=0 by using the equation f(x)=x with equal roots, but I did not find the value of a and b with the contrast coefficient.

Assuming that the first question has been solved, and the second question assumes that such mn exists, and the domain is evaluated by the definition domain of the function, and the boundaries of the range are 3m and 3n, respectively, two equations can be obtained, and the mn value can be found, if it can be found, it exists, and if it cannot be found, it does not exist.

I think it's okay to just know how to do the questions, so I'll write my thoughts here.

（1）f(x)=x²/(2x-2). 2) an=-n.(n=1,2,3...

f(x)=ax²-2ax+a+1+[1/(4a)].Please take a look at the question again, whether there are fewer conditions, a can't be found, and it's not good to do it below.

y=f(x+2) is an even function.

Then: f(x+2)=f(-x+2).

f(f (and y=f(x) is an increasing function on (0,2).

f(so:f(Question 2: Because the function is an even function, f(-x)=f(x), there is a=-1f(x)=-x 2+3

p=-(2m)^2+3=-4m^2+3

q=-(m 2+1) 2+3=-m 4-2m 2-1+3q-p=-m 4+2m 2-1=-(m 2-1) 2 is less than or equal to 0, so q is less than or equal to p

(the minimum point is) x = -a 2, the minimum value is f(x) = -a 2 4) + 3, x belongs to [-2, 2], f(x) a is brought into the solution, -(a 2 4) + 3>=a

Results: -6<=a<=2, minimum, a=-2

2.(1) f(1) = f(1 + 0) = f(1) * f(0), when x > 0, f(x) 1, both sides divided by f(1), f(0) = 1

2) Let a>0, -a<=0, f(0)=f(a+(-a))=f(a)f(-a)=1

When x>0, f(x) 1, f(a) > 1

f(a)*f(-a)=1

f(-a)>0

For any x r, there is always f(x) 0

(1) The axis of symmetry of f(x) is x=a, because it is a subtraction function on [1,2], so a <=1, g(x) in a<0 is an increasing function, and a>0 is a decreasing function, which comprehensively obtains a value range of 00,2(x-2)>0, and because f(x) is an increasing function, x>2(x-2), solve 2 Sorry, just read the wrong question.

1 [-a/2,(1-a)/2]

2 f(x)=(x-1) 2 2+3 2 When x>1, f(x) increases monotonically.

When the domain is defined as [1,m], the maximum value is obtained when x=m, and because the value range is also [1,m], m=f(m), which is substituted for m=1 2m 2-m+3 2 to obtain m=1,3(m>1).

Therefore m=3