True False questions for finding the monotonicity of a function! Function monotonicity exercise prob

For example, y=x-1 x, because 1 x is a subtraction function on (- 0) and (0,+), and because x is an increasing function on (- 0) and (0,+), the original function is an increasing function on (- 0) and (0,+).

Example 2, y= (x+1) Since u=x+1 is an increasing function at [-1,+, then y= u is an increasing function at [0,+, then the monotonicity of the composite function"If you are the same, you will increase, and if you are different, you will be reduced"It can be seen that the original function is in [-1, + is an increasing function.

The above two questions are judgments, and the following ones are proved with definitions.

For example, prove that f(x)=x+1 x is an increment function at [1,+.

Proof: Take 1 x1 x2, x1, x2 is any real number on [1,+.

f（x1）-f（x2）=x1+1/x1-x2-1/x2（x1-x2）（1-1/x1x2）

x1＜x2 ∴x1-x2＜0

x1 1, x2 1 x1x2 1 1 1 x1x21-1 x1x2 0

x1-x2）（1-1/x1x2）＜0

f（x1）-f（x2）＜0

f（x1）＜f（x2）

f(x)=x+1 x at [1,+ is an increment function.

Example: Prove that y=x 2+1 is an increment function on (0,+.

Proof: take x1, x2 (0,+ and x1 x2 0f(x1)-f(x2)=x1 2+1-x2 2-1(x1-x2)(x1+x2).

x1＞x2＞0

x1-x2＞0, x1+x2＞0

x1-x2)(x1+x2)＞0

f(x1)-f(x2) 0 f(x1) f(x2)y=x 2+1 is an increment function on (0,+).

Find the derivative of the function.

For example: (x 2).'=2x, (e x)=e, etc.

Because y'=2x, when x<0, y'< 0, so y=x 2+1 decreases monotonically on (- 0).

Obviously, when y'When x belongs to a certain range, there is always y'<0, then the function y decreases monotonically in this interval; On the contrary, when y'>0, it increases monotonically.

1.The formula f(x)=-(x-2) +4, and then draw a simple diagram, the axis of symmetry is x=2, the parabola opening is downward, and the image knows that (- 2] is the increase function. (2, + is a subtraction function.)

2.The question should be y=2x (x-1)......

3.A typical "Nike function" (or "tick function") draws a simple image (that is, in the first quadrant is a "tick", but there is no intersection with the coordinate axis, the lowest point corresponds to the abscissa is 1, (0, 1] is a single subtraction, (1, + is a single increase; The same is true in the third quadrant, except that the "hook" is rotated one hundred and eighty degrees, so (- 1] is a single increase, (is a single decrease.)

4.Are you sure the questions are complete?

1 y=(2k+1)x+b is a subtraction function on r, then 2k+1 0

k＜-1/2

2 a + b＞0

There is a-b, b-a

It is known that the function f (x) is an increasing function on r.

Then f(a) f(-b) f(b) f(-a) co-directional inequalities are added f(a) +f (b) f (-a) +f(-b).

Therefore, the A3 topic is incomplete.

4 f（x）=4x^2-mx+5

Axis of symmetry = m 8

When x (-2,+ is an increasing function, m 8 -2 , m -4 is an increasing function when x (-2,+ is an increasing function, and when x (-2) is a decreasing function m 8=-2 m=16

f（x）=4x^2+16x+5

f（1）=4+16+5=25

5 No titles.

6 Do the bad method.

Set x1 x2

f（x1）-f（x2）=（-x1）^3+1-(-x2)^3-1=-x1^3+x2^3

x2-x1)(x1 2+x1x2+x2 2) 0 f(x1) f(x2).

So f(x) is a subtractive function on r.

7 f(x)=8+2x-x^2

f（2-x^2)=8+4-2x^2-4+4x^2-x^4=g(x）g（x)=-x^4+2x^2+8

Use the derivative to find monotonicity.

July n0

Question 1 is incorrect, (-2] should read (- 2] Proof: Definition.

1.Proof: Set x10

So 1 f(x1)-1 f(x2)=[f(x2) f(x1)] [f(x1)f(x2)]>0, 1 f(x1)-1 f(x2) 0, 1 f(x1) 1 f(x2).

That is, the function y=1 f(x) is a subtraction function on the interval a, July s2

According to the steps of monotonicity definition, 1, set the interval, 2, make the difference, 3, deform, 4, and judge the completion of the symbol.

f'(x)=-2x+4 req f'(x)=0 then x=2 draw**---x (-2) 2 (2 ,+f'(x) +0 --f(x) increase greatly decreased.

So f(x) in (- 2] is an increasing function.

With derivatives··· It's very simple·· No, have you learned? It seems to be in the textbook of the second year of high school...

Choose A for this question

y=x(1-x2), when x(-1,1) y'=[（x）'（1-x^2）-x（1-x^2）']/（1-x^2）^2=（1+x^2）/（1-x^2）^2>0

The original function at (-1,1) is a monotonic increasing function.

12 Question A: Monotonous increase.

Because y' = (1+x^2) / (1-x^2)^2 > 0。

12. y = x/(1-x^2), y' = [(1-x^2) -x(-2x)]/(1-x^2)^2 = (1+x^2)/(1-x^2)^2

In (-1, 1), y'> 0, the function increases monotonically. Pick A.

Certificates: Set - x1, x2 -3, + and x x2 x1 0

y f x2 f x1 x2 squared + 6x2 x1 squared + 6x1 x2-x1 x2+x1 +6 x2-x1 x2-x1 0,x2+x1 0 y 0 The function is monotonically incremental in intervals.

2. The format of the 2 questions can be calculated.

First, find the undefined point, the stationary point, and then divide the interval. Find the first derivative of the function, and then judge the positive and negative properties of the derivative in each interval, if it is positive, it will increase monotonically in this interval, and if it is negative, it will decrease monotonically.

Ensembles are not an isolated discipline. The reason why it is placed together with functions is inextricably linked. For example, the question says that f(x) is in ......is an increasing function, then we will think of finding the monotonicity of f(x) first, finding the increasing interval, and then letting ......is a subset of the incrementing interval.

The inequality is then solved using the number axis method. There are simply too many applications of collections in functions.

In my opinion, it is not the key to do the questions, the key is to lead the students to review the common question types and common problem handling methods that we have talked about, and combine some typical and easy test questions. In terms of question types, for example, there are several methods for evaluating the function domain, such as monotonicity, discriminant method, commutation method, separation Changshu observation method, ......Common problem solutions, such as how to deal with abstract functions of the f(x) type, such as finding periods, finding symmetry axes, using monotonicity to remove f, using parity monotonicity to move to f......Wait, I think, the most important thing is these question types, methods, and then, seeing that the students have almost mastered it, ring the bell in the appropriate class, and give a more difficult question, so that it is okay. Then, about the set of papers, of course, you have to do it, and the difficulty should be moderate to ensure that the coverage is complete.

Of course, it's also good to come up with the questions yourself to ensure coverage)

When there is f(x)>0 for any x r, then f(x2) > 0, so for f(x1) f(x2)<1, both sides are multiplied by f(x2) to have the step f(x1)0".

No. To prove the monotonicity of a function on r, only the case of x>0 is given, and the case of x<0 is unknown.

If f(x) < 0 when x<0, then the next step is not true.

Landlord, when it comes to proving the monotonicity of functions, there is an implicit premise. That is, the premise of judging the monotonicity of a function is that the domain of the function must be symmetric. Without this premise, the parity of functions is not discussed.

Therefore, in this problem, we must first prove that the domain is a whole real number.

Answer: It cannot be skipped, it must be proved that there is f(x)>0 for any real number, because the problem only gives f(m+n)=f(m) f(n), monotonicity must also be solved by using this formula, and the use of division must ensure that the denominator is positive, and the inequality will not change signs. You probably get the idea?

f(x1) /f(x2)＝f(x1−x2)＜1∴f（x1）＜f（x2）

This step needs to be all positive, to give a counter-example.

2 -1 1 but 2 -1

Because, if f(x)>0 is not proved, f(x1) f(x1) f(x1) f(x2) f(x1) f(x2) f(x1) f(x2) f(x1) f(x1) f(x1) f(x1) f(x1) f(x1) f(x1) f(x1) f(x2) f(x1) f(x2)

Some of them are fixed format plus it.