A few doubts in the learning of functions, a few questions about functions

1）f(x-a)=f(x+b)

We might as well make y=x-a

Substituting the equation to get f(y)=f(y+a+b), and then replacing y with x gives f(x)=f(x+a+b).

Note: Since x and y are both independent variables, we can of course convert them to each other) For proof f(x-a+1)=f(x+b+1) we can as: y=x-1; --x=y+1

Substituting yield: f(y+1-a)=f(y+1+b);

Let x=y; Substituting gets: f(x-a+1)=f(x+b+1) since f(x)=f(x+a+b);

Then it means that without a+b, the value becomes the original value, so the period is a+b2) is the same as the first question.

3) Of course, g[g(x)]=g[-g(x)], in this case, you have to think of the whole g(x) as a variable.

1.Yes. The contents of f( ) parentheses can be added or subtracted on both sides of the equation.

2.No way. The equation is written as f(x+4)·f(x-1)=1, and if you add two parentheses at the same time, the function image will move.

3.The former is not necessarily right, and the latter is right. If g(x) is not an even function, then the former one is not true.

If you feel that what the teacher is saying is not very clear or you do not understand it well, look at the things in () as a whole, move the whole to the left or right, draw the figure, and you can see its characteristics when you draw more, whether it is periodic or symmetrical.

1, Odd functions are all central symmetric functions, symmetric centers are (0,0)2, are central symmetric functions all odd functions? No, e.g. y=x+1, is a symmetric central function, but not an odd function.

3. Are even functions periodic? No, for example, y=x 24, are all periodic functions even? No, for example, y=sinx5, are axisymmetric functions all even functions? No, e.g. y=(x-1) 2

1: Algebraic formula.

2: Scope of change. Zen guessing (also called the "definition domain" of a function.) ）

Having a solution is δ 0

When a=0, -2≠0 does not hold, and is rounded.

When a = 1, a = 1

When a=-1, x x 2=0 has a solution for [-1,1], when a=1 x x 2=0 has a solution for [-1,1], when b 4ac=a 4*a *-2=9a 0 rounded up, the value range of a is -1,1

(1) Obviously, a≠0

Let f(x)=a x +ax-2, since f(0)=-2 0, the original equation has a solution to [-1,1] when only f(1) 0 or f(-1) 0 is required.

Solutiona (-1] [1,+

2) It should be that there is always a solution on r.

Obviously, a≠0

Then make 0 solve the range of a! The result is a≠0

Because 1<=cos h<=1, 1 cosh>=0, because on the denominator, 1 cosh is not equal to 0, it can only be equal to zero.

First of all, the landlord should be aware that the power of lnx of e is equal to <> x

So the LN2 of that E is the base shirt 2

The picture below has my own method of looking at the grinding cavity.

It's not easy to answer the question, thank you].

1.Bounded.

2.Non-integrable, monotonous (the left side of the question should be an open interval) 3Accumulative.

4.If n starts at 1, then n!!It refers to 1*3*5*7*.

If n starts with 2, then n!!It refers to 2*4*6*8....

f(2)=2 (-t 2+t+2), f(3)=3 (-t 2+t+2), f(2) is f(3 2) = (3 2) (t 2+t+2) and the key is 1

t=0, f(x)=x (2), t=1, only sell f(x)=x (2).

Take x, y =0

There is f(0) = f(0) + f(0).

f(0)=0

Take x=-yf(0)=f(x)+f(-x)=0f(x)=-f(-x).

f(x) is an odd function.

g(x) is a monotonically increasing function.

The evidence is scattered.

f(x) is an odd function.

So f(x) at (negative infinity, 0) is also a subtraction function.

1 f(x) at (minus infinity, 0) is the increasing function of the single envy of the impulse sensitivity. Certification.

f(0)=f(0)+f(0)

f(0)=0

0=f(0)=f(x)+f(-x), so the number of strange towns is buried in the royal ants.

f(x) is constant on (-no dust, 0) and is single-subtracted.

So g(x) is single-incremented.