Prove with the definition that x 2 2 x 3 at infinity, 1 is an increasing function

Solution: Might as well set: - x1 x2 1

Substituting x1 and x2, f(x) = f(x2)-f(x1) = -x2 +2x2+x1 -2x1=(x1-x2)(x1+x2-2).

x1 x2 1, (x1-x2) 0,x1 x2 1, x1+x2 2,(x1+x2-2) 0

f(x)＞0

Hence there is f(x2) (fx1).

It can be concluded that the function is an increment function on - x1 x2 1. Certification.

Set x1 x2, x1, x2 (-1).

x1²+2x1+3 ）-x2²+2x2+3 ）（x2²-x1²）+2（x1-x2）

x1+x2)(x2-x1)+2(x1-x2)(x1-x2)[2-(x1+x2)]

x1＜x2x1-x2＜0

x1＜1，x2＜1

x1+x2＜2，2-（x1+x2）＞0

Then (x1-x2)[2-(x1+x2)] 0 when x1 x2,x1,x2 (-1),f(x1) f(x2), so the function is an increment.

Evidence search: f(x)=x 2+2x+1

So. f'(x)=2x+2

So. f'(x)>=0

So. 2x+2>=0

x> debunking = -1

Therefore, f(x) is an increasing function on [-1, which is poor.

y = x^2 - 2x - 3

The image is a parabola with an opening pointing upwards.

Decompose the factor. y = x-3)*(x+1)

It can be seen that the image intersects with the x-axis at x=3 and x=-1 points next to the cave.

The bottom point of the parabola is the comma x= (3-1) 2 = 1, that is, on the side of x>1, the parabola is an increasing function.

Let the unknown x1,x2,x1x1 be known, x1,x2 [-1,+ obviously have (x2-x1)>0,(x2+1)(x1+1)>0 so y2-y1>0

So functions are by definition increments.

There are two ways to do this: 1) Utilize the derivative to do this:

When x>0, f'(x)=2 x 2>0, so f(x) is monotonically increasing at (0, +infinite).

2) Definition of Utilization:

Let the circle be 0f(x1).

So the function next to Lu Xun f(x) is monotonically increasing on (0,+infinity).

Proof: For the resistant mu contains >0, there is a positive number m=1 Changxiao (+3), so that for all |x|=1/|x|-31/m-3

When x->0, (3x-1) x tends to infinity.

The original formula = (3x-1)*(1 field closed x), because x tends to be close to the Qing spinal nucleus 0, (3x-1) is close to -1, and 1 x is close to infinity, so = (3x-1)*(1 x) is close to infinity.