Prove that the function y x 1 x is an increasing function on 1, positive infinity .

Let x1,x2 (1, positive infinity), and x11,x2>1,x1*x2>11 x1*x2<11-1 x1*x2>0f(x1)-f(x2)< 0, so x is an increasing function on (1, positive infinity).

Using calculus, direct derivation.

y` = 1 - x^(-2)

When x > 1, x (-2) < 1, i.e., y >0, and y = x + 1 x is the increment function.

Definitional Method. Let x1>x2>1 then f(x1)-f(x2)=x1+1 x1-x2-1 x2=(x1-x2)[1-1 (x1*x2)].

Because x1>x2>1 so x1-x2>0 and 1-1 (x1*x2)>0, so f(x1)-f(x2)=(x1-x2)[1-1 (x1*x2)]>0

i.e. f(x1) > f(x2).

So the function is an increment function.

Hypothetical 10, because 11, 1 ab < 1, (1-1 ab) >0

It can be seen that (a-b)(1-1 ab)>0, the problem is proven.

y'=(x 2-1) x 2 When x>1, the numerator is greater than zero, so y'>0, so the function is incrementing at (1, positive infinity).

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.

Order x1> which does Shanhu Xingx2>1

f(x1)-f(x2)

x1+1/x1-x2-1/x2

x1 x2+x2-x1x2 -x1) x1x2 denominator x1x2>0

Molecule = x1x2(x1-x2)-(x1-x2)(x1x2-1)(x1-x2).

x1>x2, so x1-x2>0

x1>1, x2>1, so x1x2-1>0

So the neutron is greater than 0

So x1>x2>1, f(x1)> f(x2) are increments.

Take x1, x2, 10

That is, the function monotonically increases the rent sparrow.

Proof that the function y=x+1 x is an increasing function on (-infinity, -1): let x1<-1,x2<-1, and x10,x1<-1,x2<-1

then x1<-1<0,x2<-1<0,x1*x2>0

<-1 by x1, <-1 by x2

Get x1+x2<-2

1+x1+x2<-1

There is -(1+x1+x2)>1 (1) by x1<-1, x2<-1

Get x1+1<0

x2+1<0, then (x1+1)*(x2+1)>0

x1*x2+(x1+x2+1)>0

x1*x2>-(1+x1+x2), and -(1+x1+x2)>1 (1) get x1*x2>-(1+x1+x2)>1

Thus x1*x2-1>0

x1*x2>0

There are (x1*x2-1) (x1*x2)>0

x1*x2-1>0

There are (x2-x1)(x1*x2-1) (x1*x2)>0 to get f(x2)-f(x1)>0

i.e. f(x2) > f(x1).

by x1f (x1).

The function y=x+1 x is an increasing function on (-infinity, -1).

y'=1-1 x >0 x (-1) so increment the function.

Agree with this solution, finding a guide is the easiest.

Let x1,x2 (1, positive infinity), and x11,x2>1,x1*x2>11 x1*x2<1

1-1/x1*x2>0

f(x1)-f(x2)<0

So x is an increasing function on (1, positive infinity).

I wish you progress, and if you can't ask me again.

I'll be happy to answer your questions

Well, the boss is subtracting, and if you add it, it's obviously a subtraction function.

Using the derivative function, it's very simple!

Order 1 x1 x2

f(x2)-f(x1) = 【x2+1/x2】-【x1+1/x1】= (x2-x1) +1/x2-1/x1)= (x2-x1) -x2-x1)/(x1x2)= (x2-x1)[1 - 1/(x1x2)]= (x2-x1)(x1x2-1)/(x1x2)∵1＜x1＜x2

x2-x1＞0，x1x2-1＞0，x1x2＞0∴f(x2)-f(x1) = (x2-x1)(x1x2-1)/(x1x2)＞0

The function y=x+1 x is an increasing function on (1, positive infinity).

y'=1-1 x 2, and on (1,+, x 2>1, so 1 x 2<1, so y'>0, so the function y=x+1 x is an increasing function on (1,+.

It's so simple, because y=x is an increasing function, and y=1 x is also an increasing function on (1,+infinity), so y=x+1 x is an increasing function on (1,+infinity).

Derivative, y'=1-1 x 2, x y'>0 at (1, positive infinity), so is an increasing function.

y=x+1 x has a question and knows that x is not equal to 0

then y'=1-1 x2=(1+1 x)(1-1 x)=0, then 1+1 x=0

1-1 x=0 is solved to x=-1 or x=1, then the function y=x+1 x is in (negative infinity, -1), (1, positive infinity) is an increasing function, and (-1,0)(0,1) is a decreasing function.

In summary, the function y=x+1 x is an increasing function over (1, positive infinity).

If it is an increasing function, then x1 is less than x2, f(x1) is less than f(x2), and the two equations are subtracted to see whether it is greater than zero or less than zero.

Solution: Let -10 x1+1>0 x2+1>0(x2-x1) x2+1) (x1+1)] 0, i.e. f(x2)-f(x1)>0

The function y=2x (x+1) is an increasing function on (-1, +, if there is any clarity, we will discuss it

When x is not equal to 0, the right formula divides the top and bottom by x -1 to zero, when y is negative, and increases the positive number is the same In a word, the larger the denominator, the smaller the y.

Derivative method]: When x > 1, y = 1-1 x >0, so y = x + x 1 of the parting modulus width is an increasing function on the infinity of 1 to the potato.

y1=x1+(1 x),y2=x2-(1 cavity liquid y2) x2> hunger cry x1 x1*x2> limb 1

y2-y1) (x2-x1)=1-1 (x1*x2)>0 hence the increase function.

In the process of making a difference, the main thing is to extract the common factor and obtain (x1-x2) (1-1 x1x2) to determine the Fushan Bi number.

Or it is to seek guidance and tease like celery.