A high school function question, urgent!!

1,f(0)=f(-2+2)=f(2+2)=f(4)=1, and because the maximum value is 5, draw a diagram that obviously opens downward, the axis of symmetry is x=2, through the highest point of (2,5), and through the two points of (0,)(4,1), the first question will be. (I'm sorry, I'm a junior, and I've forgotten some formulas, so I'll do the math myself).

2. The second question can be transformed into a solution to the existing problem, which can be transformed into f(x)- 5x<=m, and g(x)=f(x)-5x, which is a new quadratic function, draw a graph, and then draw the straight line y=m. You can see it when you look at it.

3. The third question is the same as the second question, and g(x)That is, g(x)>=m is constant, and it is directly done with the condition of y=m, and it is enough to move up and down.

I can only talk about these methods, and I have to practice more by myself, otherwise the college entrance examination will be like me, and I will go to the mathematics department in college!

Let the quadratic function be f(x)=ax 2+bx+c,1), and since f(x+2)=f(2-x), the function is symmetrical with respect to x=2 and b 2a=2

2), f(0)=1, a*0 2+b*0+c=1, so c=1

3) The maximum value is obtained at the vertex, so 4a+2b+c=5

Solve three equations: a=-1, b=4, c=1, analytical: f(x)=-x 2+4x+1

f(x)<=5x+m,f(x)-5x<=m

i.e. x 2-x+1< =m

The maximum value of the quadratic function g(x)=-x 2-x+1 is: (4ac-b 2) 4a (4*( 1)*1 ( 1) 2) (-4)=5 4

So m>=5 4

f(x)>2x+m, f(x)-2x>m, i.e. x 2+2x+1>m

The axis of symmetry -b 2a=-2 (-2*1)=1 obtains the maximum value at x=1, and in the interval [1,1], the minimum value is obtained when x=-1.

m<-(1)^2+2*(-1)+1

m<-2

The main idea is to look at log2x as a whole, set it to y, then you can find a and b, and the second question is not difficult to find the range, you can do the math yourself

Define the domain as r, then mx 2-6mx+m+8 0 is constant, if m=0, then 8 0, is true.

If m is not equal to 0, mx 2-6mx+m+8 is the quadratic laughing code function Evergrande ascends to the opening so the opening is upward, m>0 and the discriminant formula is less than the grip which is equal to 036m 2-4m(m+8) 0

32m^2-32m≤0

0 so 0 m 1

The domain of y= mx -6mx+m+8 is r

When m=0, 2 2>0 satisfies the defined domain r

When m>0 =36m -4m(m+8)<=0 to get 0, when m<0 f(x)=mx -6mx+m+8, the image opening downward, there must be x, corresponding to mx -6mx+m+m+8<0

And Inner Filial Piety = 0

Therefore, m takes the first command value 0 "Shen Chun = m< = 1

1)f(-x)=|x|(-x-a), when a=0, f(-x)=-f(x), is an odd function, a!=0, non-odd and non-even.

2) x>0, f(x)=x 2-ax, because a<=0, axis of symmetry x=a 2<0, monotonically increasing.

x<0, f(x)=-x 2+ax, axis of symmetry x=a 2, opening downward, when xf(1 2), the maximum value is -1-a

When >f(a), the maximum value is 1

In conclusion, when when a", the maximum value is -1-a

1. a=0, odd function; A is not 0, not odd or even.

2. a=0, f(x) increases on r;

a<0, f(x) increases in (-infinity, a 2), (a 2, 0) decreases, and (0, + infinity) increases in a single increment.

3. Under the condition of 2, -5 2

Solution: by 0==a

a>=-a i.e., 0==a, a<=1+a, i.e., -1 21-a, or -a>1+a, i.e., a>1 2 or a<-1 2, the domain is defined as.

a=a=

There's something wrong with your title, it seems to be missing parentheses.

x=4 to obtain the minimum value, indicating that x= 4 is the axis of symmetry, because the period is 2, so the center point of symmetry is 2 units different from the axis of symmetry, and the function y=f(3 4 x)=f(-(x-3 4), f(x)--f(-x)--f(-(x-3 4), it can be seen that the original function is first symmetrically transformed in the translation transformation (the translation amount is to the right translation 3 4), grasping the function image turns out to be in x= 4 to get the minimum value, so that it is easy to draw a sketch, can be judged to be d, (There is no need to do an algebraic transformation, the first floor solution is back to mislead you) over.

Extract the root number (a 2 + b 2) from the formula. So f(x)=[under the root number(a 2+b 2)]sin(x-c). It is known that the minimum value is - under the root number (a 2 + b 2) and is obtained at x = 4, so f( 4) = root number 2 2 * (a-b) = square of both sides under the root number (a 2 + b 2) is solved.

a=-b.So the original function becomes f(x) = a(sinx-cosx) = root number 2 2*a*sin(x- 4)y=f(3 4 x) = root number 2 2 * a*sin( 2-x) = constant * cosx

You can see it, so the answer is A

There is an error in the question.

The question should be: How monotonicity is the function f(x) on (1, ?

The image of the function f(x) is symmetrical about the straight line x=1, and due to the change of the base, its image may be (01) in the shape of an eight, and the axis of symmetry is on the center line of the figure eight. From the known decreasing at (0,1), we know that a>1, so that this function increases monotonically on (1, with no maximum.

Analysis: (1) f(1) = 4, f (2) = 2 12 substitution to a-b = 2

a^2-b^2=12

obtained: a=4, b=2

2)f(x)=2^(4^x-2^x)

Let g(x)=4 x-2 x

g'(x)=4^xln4-2^xln2=2*4^xln2-2^xln2=(2*4^x-2^x)ln2

For the exponential function at x belongs to [1,2], the function is incremental, so (2*4 x-2 x)ln2 is everbright at zero.

g'(x)>0

g(x) is monotonically increasing.

So f(x) is also an increasing function.

When x=2, f(x) obtains a maximum value of 2 12

1.Bring f(x)=in[1-x 1+x] into f(x)+f(y)=f[(x+y) (1+xy)]], and get in[1-x 1+x]+in[1-y 1+y] to get in[x+y1+xy].

2.Since the condition f(x)+f(y)=f[(x+y) (1+xy)], and when x 0, f(x) 0

That's the case, then let's find f(x)-f(y)=simplify ourselves, it's OK3The third digit is based on f(x)+f(y)=f[(x+y) (1+xy)], f(x)-f(y)=???

and f(find f(x)=