Explain the function of the first year of high school I can t figure out what the teacher said .

Are you a student or a teacher?

If you are a teacher, it is recommended to start with the functions in junior high school, mainly to solve the "correspondence theory" of function definition.

If you're a student, it's a good idea to ignore those explanations, as understanding the concept of a function itself is a process.

However, most high school students do not understand functions in the end, but this does not prevent them from getting high scores in mathematics in the college entrance examination.

Is it to explain courseware or something?

Let -x>0 then x"Trembling 0

So the ridge rock f(-x)=log2(-x+1);

and because f(x) is an odd function;

So f(-x)= f(x);

So -f(x)=log2(-x+1);

Sozakura Cave is f(x)=-log2(-x+1);

Solution: (1) f(x)=ax -2ax+2+b=a(x-1) +b-a+2 ax axis of symmetry x=1, quadratic term is the number of regret beats a>0, and the image opening of the image is upward [2,3] On the right side of the axis of symmetry, f(x) monotonically increases f(x)max=f(3)=a(3-1) +b-a+2=3a+b+2 3a+b+2=5 f(x)min=f(2)=a(2-1) +b-a+2=b+2 b+2=2 synonymous ,.

Solution: f(x)=ax +(3+a)x+3,a≠0

The function image is a parabola, the axis of symmetry is x=-(3+a) (2a), and the coordinates of the vertices are (-(3+a) (2a),3-(3+a) 4a)).

f(x)=(x+1)(ax+3)

The two intersections of the function with the x-axis are x=-1 and x=-3 a, respectively

Category Discussion: 1) a 0

a) At 0 a 3, f(x) increases on [-1,4] with a single-limb plexus feast, f(x)max=f(4)=16a +4a+15=4, and obtains, 16a +4a+11=0, no solution.

b) A 3, the axis of symmetry of f(x) is in the interval [-1,0], combined with the symmetry of the function.

f(x)max=f(4)=16a +4a+15=4, no solution.

2) a 0a) when -(3+a) (2a) 4, i.e., a -1 3, the function obtains the maximum value f(x)max=3-(3+a) 4a)=4 at the vertice, and obtains the calendar silver a +10a+9=0, and the solution is a=-1 or -9

b) When -(3+a) (2a) 4, i.e., 0 a -1 3, the function increases monotonically on [-1,4] between Zheng Ran, f(x)max=f(4)=16a +4a+15=4, and there is no solution.

In summary, the question can be satisfied when a=-1 or -9.

If you have any questions, please ask for this question, thank you.

f(x)=a[x+(3+a)/2a]^2+3-(3+a)^2/4a

1.When a<0, there are three cases.

1).For Li Juna, the axis of gravity is between [-1,4], f(x)max=3-(3+a) 2 4a=4, the solution is a=-9 or a=-1, and at this time, -1<=-3+a answer hole 2a) <=4, and a<=-1, then a=-9 and a=-1 satisfy the topic;

2).When the axis of symmetry is to the left of [-1,4], i.e., -(3+a2a)<=1, then f(x)max=f(-1)=4, a has no solution.

3).When the axis of symmetry is on the right side of [-1,4], that is, -(3+a2a)>=4, then f(x)max=f(4)=4, the solution a=-11 20 satisfies the problem;

2.When a>0, there are two scenarios.

1).The axis of symmetry x=-(3+a 2a)<=1+4) 2, where f(x)max=f(4)=4, a has no solution.

2).Symmetry which axis x=-(3+a 2a)>=1+4) 2, where f(x)max=f(-1)= no solution.

In summary, there is a, and when a=-9 or a=-1 or a=-11 20 fits the topic.

Solution: f(3x+1) is defined as [-1,2]-1 3x+1 2,-2 3x 1,-(2 3) x (1 3)(1 3) x+1 (4 3).

f(x+1) is defined as [1 3,4 3].

3x+1 is greater than or equal to -1 less than or equal to 2 calculate the value of x and then calculate the value of x+1.

f(x)=2^x，f(f(x))=2^(2^x)g(x)=4^x=2^2x，g(g(x))=2^(2*2^2x)=2^(2^(2x+1))

g(g(x))>f(f(x))

Then 2 (2 (2x+1))> 2 (2 x) are monotonicity by the exponential function.

2^(2x+1)>2^x

then 2x+1>x

Hence x<1

The domain is not defined correctly.

Yes, yes. No, 1 x corresponds to 2 y, not a function.

Therefore, C. is chosen

By definition, an even function is a function that satisfies f(x)=f(-x), i.e., f(x)-f(-x)=0, and for this problem, let x<0, then -x>0

So there is f(x)-f(-x)=0

Since -x>0, you can bring f(-x) into it with the condition of the problem, and get f(x)-f(-x)=f(x)-sin(-2x)-cos(-x)=0 to solve f(x), and x<0 at this time, f(x) is the requirement of the problem.

Find: f(x)=sin(-2x)+cos(-x)=-sin2x+cosx

(1) takes the maximum value at x=3, so (3) = 2, so =3 2

2) When x<0, -x>0, so f(-x)=sin(-2x)+cos(-x)=-sin2x+cosx, and even function, so f(-x)=f(x), so f(x)=-sin2x+cosx

The monotonically increasing interval of the sinusoidal function is [k - 2, k + 2], while the increasing interval in the problem is [0, 3], and the decreasing interval is [ 3, 2] so that k=0, ( 2)= 3, so x=3 2