Prove with the definition that f x x2 increments monotonically on 0, .

Because the domain is defined as r

So let 0 tangent function.

As an example to illustrate:

The first step is to clarify the domain in which the function is defined.

Second, if the domain of the function definition is not symmetric with respect to the origin, it is a non-odd and non-even function.

Meet the definition domain with respect to origin symmetry, discuss whether it has parity or not.

Use f(-x) to calculate and simplify, and find f(-x)=f(x), which is an even function, and f(-x)=-f(x), which is an odd function, otherwise it is a non-odd and non-even function.

f(x)=tanx, the domain is defined as, so with respect to the origin symmetry, and since f(-x)=tan(-x)=-tanx=-f(x), it is proved that the tangent function is an odd function.

Secondly, let's look at the monotonicity of the tangent function, we have learned that its image is monotonically increasing in various intervals, how to prove it? First of all, it is clear that the tangent function is a periodic function with the minimum positive period, so we take (-2 ,2 ) to study. The derivative of the tangent function is 1 (cosx) 2, because cosx ≠ 0, so 1 (cosx) 2>0, so the slope is always greater than 0, thus proving that the tangent function is monotonically increasing in (-2 , 2), and from periodicity it can be deduced in the interval (-2 +2k ,2 +2k )k z, monotonically increasing on the interval, but not in the defined domain.

Proof: Set any x1

x2 (0, + and x1.)

x2 so there is f(x1

f（x2=xx（x1

x2 (x1x2 because 0 x1

x2 so x1

x20，x1

x20, so f(x1

f(x2 0, i.e. f(x1

f(x2, so the function y=x2

In x (0, + is a monotonically increasing function

The solution f(x) = (2x+4) under the root number is a monotonic increasing function on [-2,+, which proves that x1 and x2 belong to [-2,+ and x1 x2

then f(x1)-f(x2).

2x1+4）-√2x2+4）

(2x1+4)- 2x2+4)] 1

（2x1+4）-√2x2+4）]×2x1+4）+√2x2+4）/√2x1+4）+√2x2+4）]

2x1+4)) 2x2+4)Regiment) ]2x1+4)+ 2x2+4)]

2x1+4）-（2x2+4）]/2x1+4）+√2x2+4）]

2x1-2x2）]/2x1+4）+√2x2+4）]

From x1 x2 to 2x1 2x2, that is, 2x1 2x2, that is, 2x1-2x2 0

There is also x1, and x2 belongs to [-2,+ i.e. (2x1+4)+ 2x2+4) 0

Namely. （2x1-2x2）]/2x1+4）+√2x2+4）]＜0

i.e. f(x1)-f(x2) 0

That is, f(x) = under the root number (2x+4) is a monotonically increasing function on [-2, + 晌 or liter).

Take either x1, x2 (-1,1).

x=x1-x2>0

y=x2/(x2^2+1)-x1/(x1^2+1)[x2(x1^2+1)-x1(x2^2+1)]/x2^2+1)(x1^2+1)]

x2*x1^2-x1*x2^2+x2-x1)/[x2^2+1)(x1^2+1)]

x1*x2(x1-x2)+x2-x1]/[x2^2+1)(x1^2+1)]

1-x1*x2)(x2-x1) [x2 2+1)(x1 2+1)]So for the acorn celery to increase spring, such as the letter of repentance.

I see your doubts.

The basic cubic variance formula is

x³-y³=(x-y)(x²+xy+y²)

Where, y = 1* y = (1 4 + 3 4) * y = 1 4 * y + 3 4 * y

In this way, we can find that this place has formed a completely flat way.

Because xy = 2 * x * (1 2 * y).

And (1 2 *y) = 1 4 *y

So x + xy + y = x + xy + 1 4 * y + 3 4 * y

x² +2* x *(1/2 *y) +1/2 *y)²]3/4* y²

x+1/2*y)²+3/4* y²

Take x1 on (1, staring at the grandson), x2 makes x1 x2, then.

f（x1）-f（x2）=2x1^2-4x1-2x2^2+4x22(x1^2-x2^2)-4(x1-x2)2[(x1+x2)(x1-x2)]-4(x1-x2)2[(x1+x2)(x1-x2)-2(x1-x2)]2[(x1-x2)(x1+x2-2)]

x1 x2, Kailian x1-x2 0

x1，x2∈（1，＋∞x1＞x2＞1，∴x1+x2＞2，∴x1+x2-2＞0

x1-x2)(x1+x2-2)]＞0，∴2[(x1-x2)(x1+x2-2)]＞0

f(x1)-f(x2) 0,f(x1)>f(x2)f(x)=2x -4x is an increment function on (1.

Qiyuan f(x)=2x -4x at (1, monotonically increasing on it.

You try it with a function demodulation!

Let x1>x2 2, then:

f(x1) f(x2).

x1) (4 x1)] x2) (4 x2)](x1 x2) [4(x2 x1)] x1x2)[(x1 x2) (x1x2)] x1x2 4) because: backup x1>x2 2, then:

x1 x2> Slippery Hunger 0, x1x2>4, then:

f(x1) f(x2)>0, i.e.:

f(x1)>f(x2)

So the function f(x)=x(4 x) is incremented on [2,.

f(x)=x+1 x, this function is very important, and it must be clearly understood. Ha, that's a reminder.

Proof: Let 11,x1-x2<0,x1-x2)(x1x2-1) x1x2<0, i.e., f(x1)-f(x2)<0

So f(x)=x+1 x is monotonically increasing on (1.

In (1, take x1, x2 makes x1 x2, then.

f（x1）-f（x2）=2x1^2-4x1-2x2^2+4x22(x1^2-x2^2)-4(x1-x2)2[(x1+x2)(x1-x2)]-4(x1-x2)2[(x1+x2)(x1-x2)-2(x1-x2)]2[(x1-x2)(x1+x2-2)]

x1＞x2,∴x1-x2＞0

x1,x2 (1, x1 x2 1, x1+x2 calendar size 2, x1+x2-2 0

x1-x2)(x1+x2-2)]＞0,∴2[(x1-x2)(x1+x2-2)]＞0

f(x1)-f(x2) 0,f(x1)> band f(x2)f(x)=2x -4x is an increment function on (1).

f(x)=2x -4x in monotonically increasing increments on (1.