Judge the monotonicity of the function y x 2 x 1 and prove

There are two ways to solve this problem, the definition method and the differential method.

Solution 1 (Differential Method).

From y = (x+2) (x+1), we know that the domain of the function is x (-1) (1,+).

The first derivative of y = (x+2) (x+1) is .

y! = -[1/(x+1)2] ＜0

i.e. the function y = (x+2) (x+1), which defines x (-1) (1,+ on.

y!0 function y = (x+2) (x+1), define x (-1) (1,+ on.

is a subtraction function solution two (definition).

x1 and x2 are any two defined on (- 1) (1,+.

number, and x1 < x2

Knows by y (x+2) (x+1).

y2-y1＝[(x2+2)/( x2+1)]－x1+2)/( x1+1)]

x1 －x2）/( x1+1)×(x2+1)

y2－y1＝（x1 －x2）/( x1+1)×(x2+1)

When x1, x2 is defined at (-1), x1 < x2.

x1 －x2＜0, (x1+1)＜0, (x2+1)＜0

y2－y1＝[（x1 －x2）/( x1+1)×(x2+1)]＜0

y2 y1 i.e. when x1 and x2 are defined in (- 1), x1 < x2, y2 y1

y (x+2) (x+1) where (- 1) is a subtraction function.

When x1, x2 is defined at (-1), x1 < x2.

x1 －x2＜0, (x1+1)＞0, (x2+1) ＞0

y2－y1＝[（x1 －x2）/( x1+1)×(x2+1)]＜0

y2 y1 i.e. when x1 and x2 are defined in (- 1), x1 < x2, y2 y1

y (x+2) (x+1) where (- 1) is a subtraction function.

Synthesis , learned.

The function y (x+2) (x+1) is defined as subtracting on (- 1) (1,+).

There are three methods, the first of which is the definition method.

Let any x1 and x2 belong to r and ≠-1, and x1 is scored, the result is obtained.

The second, deriving, is upstairs.

The third is observation.

Original = 1+[1 (x+1)].

The larger the x, the smaller the fraction after it, and the smaller the overall y, so it is minus, but not on r, when x≠-1.

Monotonically decreasing for the following reasons: y=x+2 x+1=(x+1+1) x+1=1+1 (x+1), the larger the x, the smaller 1 (x+1) is, the smaller the y. So it's monotonically decreasing.

Dear, do you understand, I hope it will be useful to you!

Enlightenment: y'=1-2/(x^2)

infinity, - root number 2), (root number 2, + infinity) is incremental.

root number 2, + root number 2) is minus.

The domain of the original function is x≠0, let x be positive, and take x from the mean inequality x+2 x 2 2 to get the minimum value when x takes 2, and we get (0, 2] monotonically decreasing, ( 2,+ monotonically increasing, (-2,0) monotonically decreasing and (-2] monotonically increasing by the properties of the double-hook function.

Proof: Let y=1 u(x) u(x)=x 2-1 x 2-1≠0 x≠1 or -1

y=1 u(x) is a subtraction function.

u(x) is the subtraction function at (-1,-1) and (-1,0); In [0,1) and (1,+infinity) is an increasing function.

So y=1 (x 2-1) is an increasing function in (-infinity, -1) and (-1,0); In Paga [0,1) and (1,+infinity) is a subtractive function.

Ren Cong Zhengliang takes x1 less than x2 to belong to this interval, f(x1)-f(x2)(x1-x2)+(1 x1-1 x2).

x1-x2)+(x2-x1)/x1x2

x1-x2) (1-1 clear x1x2).

x1-x2)(x1x2-1) Infiltration width x1x1 Because x1 is less than x2, x1-x2 is less than 0

Because x1 x2 is greater than 1, so x1x2-1 is greater than 0, so f(x1)-f(x2) is less than 0

So f(x1) is less than f(x2).

So it's an increment function.

f(x)=1/x

Defining the field x is not equal to 0

Let a>b>0

f(a)-f(b)=1 a-1 b=(b-a) (ab)a>0,b>0, so the denominator is greater than 0

a> excavates b, b-a<0, and the small molecule is blocked at 0

So a>b>0.

f(a)0, f(x) is the subtractive function of Pei Sanhu. Similarly, af(a) > f(b).

So when x<0, f(x) is also a subtraction function.

So x>0 and x<0, y=1 x are both subtractive functions.

When x strikes Stuffy 0.

y=x+1/x≥2√[x*1/x

When x = 1 x

, i.e., x=1).

Therefore x decreases at (0,1) and increases at [0,+infinity);

When x 0, y=

x+1/x≤-2√[x*1/x

When x = 1 x

When taking the cover to make an equal sign, that is, the cheating x=-1).

Therefore x is increasing at (infinity, -1) and decreasing at [-1, 0).

The monotonicity of the function y=1 x, which is a subtraction function on (0,+) and a subtraction function on (- 0).

Prove that if x1 and x2 belong to (0,+ and x1 x2 then f(x1)-f(x2).

1/（x1）-1/(x2)

x2）/（x2）（x1）-（x1）/（x1）(x2)=（x2-x1）/（x2）（x1）

By x1, x2 belongs to (0, + and x1 x2

i.e. x1 0, x2 0, x2-x1 0

i.e. (x2-x1) (x2+1)(x1+1) 0, i.e. f(x1)-f(x2) 0

Therefore the function y=1 x is a subtraction function at (0,+).

In the same way, we can see that the function y=1 x is a subtraction function on (- 0).

For y=x+1 x, we derive:

y '=1-1 x =(x -1) x x >=0, so when x>1 or x<-1(x -1) > 0,y'>0, the original function is monotonically increased when -1< x<1,,y'<0, the original function decreases monotonically.

The original function can be deformed into a cover beam: y=x+1 x.=-x-1 x), where qing-x>0, according to the fundamental inequality we know that the function y=-(x-1 x) has vertices (-1, -2) on (- 0).

Take x11 to get f(x1).

f(x2) is a subtractive function on (-1,0).

In summary, y=x +1 x is called an increasing function on (- 1) and a decreasing function on (-1,0).

You can use the definition: 0 is not defined, divided into positive numbers, negative numbers are discussed, and x1 and x2 sets of monotonic definitions can be set (f(x1)-f(x2)). You can also sit by the derivative. Monotonically decreasing over the corresponding interval.

Since x≠0, the two interval analysis is set in the same interval and there is x1 in the same interval, because x1 and x2 have the same sign in the same interval, so x1x2 is a positive number in the formula x11 x2<1 x1, which shows that the function is a subtraction function in the above two intervals.

Monotonically incremental. Finding the derivative (x=0), the derivative is always at zero.

The function increases at (negative infinity, -1) and (-1, 0).

Subtract at (0,1) and (1,positive infinity).

Since the function y=1 (x 2-1) is the opposite of the monotonicity of the function f(x)=x 2-1, it is sufficient to calculate the monotonicity of the function f(x).

Proof: Let y=1 u(x) u(x)=x 2-1 x 2-1≠0 x≠1 or -1

y=1 u(x) is a subtraction function, and u(x) is a subtraction function in (-infinity, -1) and (-1,0); In [0,1) and (1,+infinity) is an increasing function.

So y=1 (x 2-1) is an increasing function in (-infinity, -1) and (-1,0); In [0,1) and (1,+infinity) is a subtractive function.

Bring in x = -x

y(-x) = 1 ((-x) 2 - 1) = 1 (x 2-1).

y(x) conforms to the property of an even function.