Monotonicity of functions "beginner and middle-order tutoring"

The monotonicity of a function may also be given in what form?

1, give directly;

such as function in Interval \ ([a,b]\) monotonically increment;

2, the definition of the given;

If the function is given \ (f (x) \) satisfies on the interval \ (d\) \ (\forall x_1,x_2\in d,x_1<x_2,f (x_1) <f (x_2) \),

means that the function \ (f (x) \) is monotonically incremented on the interval \ (d\) ;

3, in the form of the definition of deformation given;

As monotonically increasing in the form of an accumulation:

As given function \ (f (x) \) satisfies on interval \ (d\) \ (\forall x_1,x_2\in D, (x_1-x_2) \cdot (f (x_1)-F (x_2)) <0\)

means that the function \ (f (x) \) is monotonically decreasing on the interval \ (d\) ;

4, in the form of the definition of deformation given;

As the monotone increment is given in the form of a commercial type:

As given function \ (f (x) \) satisfies on interval \ (d\) \ (\forall x_1,x_2\in d,\cfrac{f (x_1)-F (x_2)}{x_1-x_2}>0\)

means that the function \ (f (x) \) is monotonically incremented on the interval \ (d\) ;

5, the "monotonous + odd" in a comprehensive form given;

As given function \ (f (x) \) satisfies on interval \ (d\) :\ (\cfrac{f (x_1) + f (x_2)}{x_1+x_2}>0\), and function \ (f (x) \) is an odd function,

( f (-x_2) =f (x_2) \), substitution gets \ (\cfrac{f (x_1)-F (-x_2)}{x_1-(-x_2)}>0\),

Re-order \ (-x_2=x_3\), i.e. \ (\cfrac{f (x_1)-F (x_3)}{x_1-x_3}>0\),

That is, the function \ (f (x) \) is monotonically increasing on the interval \ (d\) ;

6, given in the form of an image, (give the \ (f (x) \) image or \ (f ' (x) \) the image, to read the slope)

For example, the \ (f (x) \) image, need to interpret the image, give the \ (f ' (x) \) the image, will pass \ (f ' (x) \) the positive and negative interpretation of monotonicity;

7, the function Monotonicity property application;

Conclusion ①: function \ (f (x), g (x) \) is the increment (decrement) function, then \ (f (x) +g (x) \) is the increment (decrement) function;

Note that this is not judged by the "same-increment-subtraction" of the compound function, but by the definition of monotonicity.

Conclusion ②: The known function \ (f (x), g (x) \) is the increment (decrement) function and is known \ (f (x) >0,g (x) >0\), then there is \ (f (x) \cdot g (x) \) is the increment (subtract) function;

The known function \ (f (x), g (x) \) is the increment (decrement) function, and also known as \ (f (x) <0,g (x) <0\), then there is a \ (f (x) \cdot g (x) \) is a minus (increase) function;

8, the monotone is given in the form of compound function;

\ (\fbox{case 1}\) (2017 Fengxiang Secondary School senior Science Second monthly exam the 9th question)
If the function \ (f (x) =log_a^\;(6-ax) \) is reduced on \ ([0,2]\) , then the value range of the real number \ (a\) is ""

A,\ ([3,+\infty) \) \ (\hspace{2cm}\) B,\ ((0,1) \) \ (\hspace{2cm}\) C,\ (( 1,3]\) \ (\hspace{2cm}\) D, \ ((1,3) \)

Analysis: Make \ (g (x) =6-ax\), such as this type of topic to consider monotonicity, but also to consider the definition of the domain.

From the topic know must have \ (a>0\), therefore function \ (g (x) \) monotonically decrement, consider defining the domain as long as the minimum value \ (g (2) >0\) can be,

Consider the external function must be an increment function, so \ (a>1\),

Combine \ (g (2) >0\), Solution (1<a<3\), so choose D.

9. Monotone is given in the form of piecewise function

\ (\fbox{case 0}\) (The monotonicity of the known piecewise function, the value range of the parameter is obtained)

Known \ (a>0\), function \ (f (x) \) satisfies \ (f (x) =\begin{cases} (3-a) x-3 &x\leq 7 \ a^{x-6} &x>7 \end {cases}\), the function \ (f (x) \) is monotonically incrementing on \ (r\) , and the range of values for \ (a\) is obtained.

Analysis: By the title,\ (\begin{cases} &3-a>0 \ &a>1 \ & (3-a) 7-3\leq a^{7-6}\end{cases}\);

That is \ (\begin{cases}&a<3 \ &a>1 \ &a\ge \cfrac{9}{4}\end{cases}\)

Solution:\ (a\in[\cfrac{9}{4},3) \);

10. The monotone is given in the form of the assignment method;

If the function defined on \ (r\) \ (f (x) \) satisfies \ (f (x+y) =f (x) +f (y) \), and \ ( x >0\) ,\ (f (x) <0\) to determine the monotonic function.

Analysis: Make \ (x_1> x_2\), then \ (x_1-x_2>0\), therefore \ (f (x_1-x_2) <0\),

Then there are $ f (x_1) = f (x_1-x_2+x_2) = f (x_1-x_2) +f (x_2) < F (x_2) $,

Therefore, the function \ (f (x) \) is monotonically decreasing on \ (r\) .

11, given in the form of a derivative,

such as function in Interval \ ([a,b]\) satisfies \ (f ' (x) \ge0\)(only at a limited number of points make \ (f ' (x) =0\))

12, given in the form of a product function,

such as \ ((x-1) \cdot F ' (x) >0\), you know \ (\left\{\begin{array}{l}{x-1>0}\\{f ' (x) >0}\end{array}\right.\ ) or \ (\left\{\begin{array}{l}{x-1<0}\\{f ' (x) <0}\end{array}\right.\)

That is, when \ (x>1\) ,\ (f ' (x) >0\), that is, function \ (f (x) \) on the interval \ ((1,+\infty) \) monotonically increment;

When \ (x<1\) ,\ (f ' (x) <0\), the function \ (f (x) \) is monotonically decreasing on the interval \ ((-\infty,1) \) ;

Similarly, you can understand the expression \ ((x^2-3x+2) \cdot F ' (x) >0\).

13, given the whole or part of the guiding function (more difficult),

For example, when the problem is given when \ (x>0\) satisfies the condition \ (XF ' (x)-F (x) <0\), it tells us that we need to construct a new function, and we can know the monotonicity of the new function.

Analysis: Construction \ (g (x) =\cfrac{f (x)}{x}\), when \ (x>0\) ,

Then \ (g ' (x) =\cfrac{xf ' (x)-F (x)}{x^2}<0\),

That is, the new function \ (g (x) \) is monotonically decreasing on the interval \ ((0,+\infty) \) .

• function monotonically increment or decrement five kinds of representative form: gradually increase type, gradually reduce type, constant not variant, first slow after fast type, first fast after slow type.

Second, the code example analysis:

"Example 01"\ (f (x) \) is even function, when \ (x\in (-\infty,0) \) ,\ (f (x) +xf ' (x) <0\) set up, compare \ (2f (2), 3f (3) , the size of 5f (5) \) .

Analysis: Construction \ (g (x) =x\cdot f (x) \),\ (g (x) \) is an odd function, when \ (x\in (-\infty,0) \) ,\ (f (x) +xf ' (x) <0\ ) , then \ (g ' (x) =f (x) +xf ' (x) <0\), so the monotone and parity is known \ (g (x) \) on \ ((0,+\infty) \) on the monotonically decreasing. The size comparison is easy.

"Example 02" known function \ (f (x) =x^3-2ax+1\) on interval \ ([2,5]\) \ (\underline{monotonically incrementing}\), parameter \ (a\) range of values to be taken.

Analysis:\ (f ' (x) \ge 0\) is established on the interval \ ([2,5]\) ,

That \ (3x^2-2a\ge 0\) is established on the interval \ ([2,5]\) ,

The separation parameter is obtained,\ (2a\leq 3x^2\) is established on the interval \ ([2,5]\) ,

i.e. \ (2a\leq [3x^2]_{min}=12\),

That is \ (A\leq 6\).

\ (\fbox{case 3}\) (constructor + size comparison)

(\cdot\) (Henan Pingdingshan) known \ (f (x) \) is defined on the \ (((0,+\infty) \) function, to any two unequal positive numbers \ (x_1,x_2\) , have \ (\cfrac{x_2f (x_1)-x_1f (x_2)}{x_1-x_2}>0\), Kee \ (a=\cfrac{f (3^{0.2})}{3^{0.2}}\) ,\ (b=\cfrac{f (0.3^2)}{0.3^2}\),\ (c=\cfrac{f (log_25)}{log_25}\), then ()

A. \ (a<b<c\) \ (\hspace{2cm}\) B.\ (b<a<c\) \ (\hspace{2cm}\) A.\ ( c<a<b\) \ (\hspace{2cm}\) A.\ (c<b<a\)

Analysis: Note the structure of \ (a,b,c\) , guessed by the topic: the function to be constructed is \ (g (x) =\cfrac{f (x)}{x}\), then whether it is correct, the following is done to verify.

(0<x_1<x_2\), the equivalent form defined by Monotonicity is available,

\ (\cfrac{g (x_1)-G (x_2)}{x_1-x_2}=\cfrac{\cfrac{f (x_1)}{x_1}-\cfrac{f (x_2)}{x_2}}{x_1-x_2}=\cfrac{x_2f (x_1)-x_ 1f (x_2)}{x_1x_2 (x_1-x_2)}\)

By topic, to any two unequal positive numbers \ (x_1,x_2\), there are \ (\cfrac{x_2f (x_1)-x_1f (x_2)}{x_1-x_2}>0\),

It is known that \ (\cfrac{g (x_1)-G (x_2)}{x_1-x_2}>0\), that is, function \ (g (x) =\cfrac{f (x)}{x}\) is monotonically increasing,

So the topic needs us to compare \ (g (3^{0.2}) \),\ (g (0.3^2) \),\ (g (log_25) \) the size of the three, only need to compare the size of the argument can be;

Because \ (1=3^0<3^{0.2}<3^{0.5}=\sqrt{3}<2\),\ (0<0.3^2=0.09<1\),\ (log_25>log _24=2\),

therefore \ (g (0.3^2) <g (3^{0.2}) <g (log_25) \), i.e. \ (b<a<c\).

Monotonicity of functions "beginner and middle-order tutoring"