Senior 1 Function Questions In Depth Study Functional questions for Senior 1

The analytic formula of the function is f(x)=3x-1, using the matching method: f(x+1)=3x+2=3(x+1)-1, that is, f(x)=3x-1, it can also be changed, so that x+1=t, then x=t-1, f(t)=3(t-1)+2=3t-1, that is, f(x)=3x-1, the list method means that the ellipsis is missing. Three points, both front and back.

The definition domain is the value range of x, not the value range of x+1, and the definition domain is for x itself. The range of x+1 represents the translation of the range of the domain, i.e. the overall translation to the left by 1 unit.

The set of function values is called a range. **It's not clear, you can call me.

Congratulations on your correct answer!!

y=f(x+1)=3x+2=3(x+1)-1, i.e., y=f(x+1)=3(x+1)-1

Defines a domain, which in mathematics can be thought of as a set of all the input values of a function. So it's the range of x values. The set of independent variables -3, -2, -1, 0, 1, 2, 3, and 4 is called the definition domain. The set of function values is called a range!

The definition field is the range of values of x.

I don't know what that x+1 set is called...

The set of function values is called a range.

Hello, f(x)=3(x-1)+2

The so-called definition field is the range of values of x. x-1 is just one of the expressions.

The range of x is the domain of definition, and the range of y, i.e., f(x) corresponding to x, is the range of values.

This kind of problem is about the definition domain of the composite function, the known function f(u), and u=h(x), the definition domain is the range of the value of the independent variable x that makes the function meaningful, for the composite function must pay attention to the hierarchy, figuratively, f is the parent function, h is the child function, first of all, h(x) is meaningful. That is, the value range of x is the definition range of u, and the value range of u is the definition domain of the parent function, that is, the value range of the child function.

For a function, the domain is the range of values of the arguments that make the function relationship meaningful.

Example (1) The domain of the function f(x) is known to be [0,1), and the functions f(x+1) and (x2) are found

the domain of the definition; The domain of the function f(x) is [0,1], which requires the domain of the function f(x+1).

That is, the definition domain equivalent to f(u) is [0,1) u=x+1, that is, the definition domain of the subfunction u=x+1.

0<=u<1==>0<=x+1<1==>-1<=x<0

The domain of the function f(x+1) is defined as [-1,0).

The same goes for 0<=x 2<1==>0<=x<1

The domain of the function f(x 2) is defined as [0,1);

A defined field is the range of values for an independent variable.

In this problem, it is the range of x+1 values.

The set of function values is called a range.

In the problem, x can be seen as a parameter of the independent variable x+1.

The range of values of x is called the range of values of parameters.

Solution: Set a==

Since a intersects b = b, the solution of x 2 + ax + a 2-12 = 0 is equal to -2 or 4x = -2 substitution, 4-2a + a 2-12 = 0, and a = -2 or 4.

Check it and find that a=-2 has two answers, rounded off, a=4.

x=4 substitution, 16+4a+a 2-12=0, a=-2 have two answers, round.

So, a belongs.

f（x）=ax²+bx+c

g（x）=-bx

The first problem is to prove that there are two intersections, at which the positions of the two equations are the same, and x and y are also the same. So we can assume: f(x)=g(x), and use this equation to prove that this equation has two different solutions, and we can get the x values of two different intersections.

1) Because f(1)=0, a+b+c=0

Let f(x)=g(x).

ax²+2bx+c=0

4b²-4ac

4(-a-c)²-4ac

4a²+4ac+4c²

a²+2ac+c²)+a²+2ac+c²)+2a²+2c²)

2(a+c)²+2(a²+c²)

So: δ 0, ax +2bx+c=0 will definitely give two different solutions x1 and x2

That is, there must be two different x1 and x2 so that f(x)=g(x) is true, and f(x) and g(x) must have two different intersections.

2) The calculation process will be more complicated, I will tell you the method.

f（x）=f（x）-g（x）

ax²+2bx+c

ax²+2bx-a-b

It needs to be categorized and discussed, and it may be easier for you to draw a picture and analyze it:

First, determine the middle line of this parabola: x=-b a

The midline gives the maximum value of the whole equation: y=(b -a -ab) a

When a 0, the parabolic opening is upwards and has a minimum.

When x=-b a is in [2,3], y=(b -a -ab) a is the minimum value of 9, and when x=-b a is in [2,5 2], f(3)=21

When x=-b a is in [5 2,3], f(2)=21

When x=-b a2, f(2)=9, f(3)=21

When x=-b a3, f(2)=21, f(3)=9

In the same way, we will discuss the case where a 0 and y=(b -a -ab) a is the maximum. In the end, depending on which one meets the situation, it must be calculated that there are many things that do not meet, such as the value of a can eliminate a lot.

It's easy to draw the picture, but the calculation process is complicated.

If you don't know how to do it, don't write it down, is it for the reward?

The first two floors of the second question were wrong.

You draw the image and you're done...

Let g(x)=(m-3m+2)x +2(m-1)x+5=(m-1)(m-2)x ++2(m-1)x+5>0

1) The function f(x) defines the domain as r, for example, m=1, then g(x)=5, f(x)=lg(5).

If m=2, then g(x)=2x+5 is not as large as 0, which is not true.

For example, 12,(m-1)(m-2)>0,g(x) is a quadratic function with the opening upward, g(x)=(m-1)(m-2)x ++2(m-1)x+5

m-1)(m-2)[x²+2x/(m-2)+1/(m-2)²]5-(m-1)/(m-2)

m-1)(m-2)[x+1/(m-2)]^2+5-(m-1)/(m-2)

5-(m-1)/(m-2)

Let g(x) >0, then 5-(m-1) (m-2)>0, and get m<2 or m>9 4, since m<1 or m>2, so m<1 or m>9 4 satisfies the f(x) definition domain r, and f(x)>=lg[5-(m-1) (m-2)].

To sum up:

If the function f(x) defines the domain r, it is satisfied only if m<=1 or m>9 4, and f(x)>=lg[5-(m-1) (m-2)].

2) The range of the function f(x) is r, and the value range of the function is of course r.

(1) [(m -3m + 2) x +2 (m -1) x + 5 x r constant formation m -3m + 2 = 0 m = 1 or 2

m=1 is true, m=2 is not rounded.

m²-3m+2>0 △<0

So m>9 4 or m<1 or m=1

2) The requirements are relatively small.

So as long as the defined domain is met.

1≤m≤9/4

Why do you always have to make up some questions deliberately for the sake of the exam?

Quadratic functions. Sine function. Exponential functions.

Logarithmic functions.

A ( 4, 2 ), sina > cosa, so is negative. For example, a sine of 60 degrees is greater than a cosine.

1. x² -1 2.From (f(a)+f(b)) pants refers to silver(a+b) 0 and f(x) is a singular function, f(x) is the function of increasing Hu Yan (substituting a=x1 b= -x2 to make fun of Zheng Ke) to use monotonicity.

It is known that the image of the function y f(x) is symmetrical with respect to the y axis and satisfies f(x 2) ax 2 (a 3) x (a 2).

1. Find the analytic expression of the function f(x).

2. Discuss the number of zeros of f(x) b(b r).

3. If f(x) b has three zeros, the function h(x) x 2+2x+c xIf any x [b, h(x) 0 is constant, try to find the range of values of the real number c.

Did you say that the square was added here? If yes, I will solve the problem according to this understanding.

Analysis: The image of the function y f(x) is known to be symmetric with respect to the y axis, which means that the function is even.

Solution: Let t=x-2, then x=t+2, f(t)=a*(t+2) 2-(a-3)*(t+2)+(a-2), f(-t)=a*(-t+2) 2--(a-3)*(t+2)+(a-2), f(t)=f(-t) can be solved to get a=-1

f(t)=-(t+2) 2+4*(t+2)-3, f(t)=1-t 2

That is, the analytic formula of the function f(x) is: f(x)=1-x 2

The analytic formula is found out, and the rest is very easy, because the mathematical symbols are not easy to play, I will not continue to do it, if you can't do it, you can continue to ask me.