High School Functions 5, High School Functions

Your scope is too big, high school functions for several semesters, a variety of different functions, such as referring to functions, power functions, trigonometric functions, light trigonometric functions There is a whole book, you just ask high school functions, but also explain specifically? It's too difficult, if you really want to understand, the words typed, the words typed out are enough to make a set of books, I really can't help, if you really don't understand, it is recommended that you find a professional teacher, pay tuition and let him tell you, or find professional books.

No, there aren't many functions in high school.

Proportional functions, primary functions, inverse proportional functions, logarithmic functions, exponential functions, quadratic functions, constant functions.

Just take a look at the above functions.

Dots (3,4) are on the image of the letter and silver enlightenment.

4=3a+1/(3+b) ①

f'x=a-1/(x+b)^2

From the tangent equation at the point (3, fx) is y=4, the tangent slope is 0, so f'(3)=a-1 (3+b) 2=0 is obtained by the two-formula solution.

a=1, b=-2 or a=16 9, b=-15 4 take the previous value as an example.

fx=x+1 (x-2).

By fx=x+1 (x-2)=x-2+1 (x-2)+2 when x>2, fx 2 [x-2+1 (x-2)]+2=4, and if and only if x=3 there is a minimum value of 4

When x<2, fx -2 [x-2+1 (x-2)]+2=0, and if and only if x=1 there is a maximum value of 0

Therefore, it is assumed that the center of fx symmetry is (2,2).

Set the point of travel (m,n) on the FX image, and its symmetry point about (2,2) is (4-m,4-n).

then n=m+1 (m-2).

It is easy to prove that f(4-m) = 4-n

Therefore, the FX image is a centrally symmetrical graph, and its center of symmetry is (2,2).

Since f(x) is a dual pure equilibrium function, the vertical doing is true for any real number x with f(-x)=f(x), and similarly, g(-x)= g(x) is true for any real number x.

Since g(x)=f(x-1), it is obtained by f(x+1)=f(-x-1)=g(-x)=-g(x)= f(x-1).

f(x+2)=f[(x+1)+1]= f[(x+1)-1]= f(x), then f(x+4)=f[(x+2)+2]= f(x+2)= f(x), therefore, f(x) is a periodic function with period 4, then f(2013)=f(1+4*503)=f(1), take x=0 in f(x+1)= f(x-1) to get f(1)= f(-1)= f(1) , So f(1)=0 , so f(2013)=0 . Pick B

g(x) is the odd function on r, then g(0)=0 then f(1)=0

Since f(x) is an even function and g(x) is an odd function, then g(-x)=-g(x) because g(-x)=f(-x-1) (replace x in the condition with -x) and f(x) is an even function, then f(-x-1)=f(x+1).

So f(x+1)+f(x-1)=0

Then you Yu f(2013)+f(2011)=0, and because high school is f(2011)+f(2009)=0

So, f(2009) = f(2013).

Similarly, f(2013)=f(2009)=f(2005)=Is it right to choose b from f(1)=0?

Hello: I think gx is an odd function, so f(x-1)+f(-x-1)=0 because fx is an even function, so f(x-1) envy + f(macro key x+1) pie J = 0 because the odd function always has g(0) = 0, so f(-1) = 0 because f(x-1) + f(x+1) = 0, so f(1) = 0 f(3) = 0... So the odd f-function is equal to 0, so f(2013)=0 choose b

The answer is b, the easiest way is to take a loss and set up a special function, which can be set to f(x)=0, g(x)=0, they are all in line with the topic, so f(2013)=0

Needless to say, if that value can be found, it must be 0

If it's a function like this, it's a composite function, so if you draw an image, you can see that it's urinating.

1: When x=y=0, f(0)=0 can be obtained from f(x+y)=f(x)+f(y).

2: Take y=-x, substitute f(x+y)=f(x)+f(y) then f(0)=0=f(x)+f(-x) so f(x)=-f(-x) so it is an odd function.

3: Because it is an odd function, f(x squared - 2x) - f(x) = f(x squared - 2x) + f(-x) = f(x squared - 2x-x) > = 8

8=4f(1)=4f(1)=f(1)+f(1)+f(1)+f(1)=f(4)

So f(x 2-3x)>=f(4).

When x+y>0 is taken, f(x+y) = f(x)+f(y)<0

i.e. f(x)<-f(f)=f(-y) at x>-y

The function f(x) is a monotonically decreasing function.

And because f(x 2-3x)>=f(4).

So x 2-3x < = 4

Solution: -1=

After emanating f(-x)= f(x) from the odd function definition, we get c=0, so f(x))=ax 2+1) bx

Then from f(1)=2,f(2)<3, we get a=1, b=1, c=0, so f(x)=(x 2+1) x This is commonly known as the nike function.

Attribute number accompaniment: Define the domain {x|x<>0}, the range (-2 [2,+ is an increment on (-1 ,[1,+.

On (0, Zhengbi Qin1), (1, 0) is a subtractive function.

f(x)=(ax^2+1)/(bx+c)

Because f(x) is a singular function.

f(-x)=-f(x)

f(-x)=(ax^2+1)/(bx+c)f(x)=-ax^2+1)/(bx+c)

Molecularly ax 2 + 1 = ax 2 + 1

The first part of the hidden pin is bx+c=bx-c

c=0 f(1)=2

So a+1=2b

a=2b-1

f(2)<3

4a+1)/2b<3

If b>0

4a+1<6b substitute a=2b-1.

2b<3

b<3/2

b=1 a=1

If b>0

b>3/2

Not true, so a=1

b=1 c=0

1.Let 1 x + 1 = t, then x = 1 t -1, i.e. f(t) = 1 t -1, i.e. f(x) = 1 x -1

Let x-2=t, then f(t)+2f(-t)=2t+4, and then swap t and -t to get 2f(-t)+4f(t)=8-4t, so 3f(t)=4-6t, f(x)=4 3-2x

2.Let t=x+1=t, the domain of t is [ 5, 10], and the domain of the two functions is the same, so the domain of x is [1+ 5,1+ 10].

3.I don't understand.

4.Let x-1=t, x=t+1, then f(t)=t +2t, f(x)=x =2x

5.-1 belongs to the defined domain x less than 2, which is directly brought into the analytic formula f(x)=2x=-2

6.Let t=x+1 x, x +1 x =(x+1 x) -2=t -2, so f(x)=x -2

x and 1 x are swapped to give f(1 x) + 2f(x) = 3 x, and f(x) is subtracted to get f(x) = 2 x-x

These are the basics of functions, no matter what the unknown x in the definition field is replaced with, it is the same drop, that is to say, f(x) and f(t) and f(m) are all a function, but we are used to use f(x) to represent it, the landlord must pay attention to this.

1 f(x)=1/x²-1

f(x)=4/3-2x

2 [One plus root number five, one plus root number ten].

3 I didn't understand what was described.

4 f(x)=x²+2x

5 Do it yourself, -2

6 f(x)=x²-2

f(x)=2/x-x

I'm dizzy. These questions must be done on your own, as these are basic. The method is equivalent substitution.

It is easy to know that w=-2, and |ab|=cos2x+2sinx=3 2 --2(sinx --1 search round 2) 2,|ab|At maximum, x = 6

The derivative of f(x) is -2sin2x, and the derivative of g(x) is -2cosx, and when x=6, the derivatives of the two functions are equal, i.e., tangents of the same slope at ab, and thus the original proposition is proved.