Proof that the function f x 1 x x decrements monotonically at 0,1 .

Method 1: To prove that f(x)=1 x+x decreases monotonically at (0,1), you can set 0 as long as you prove f(x1)-f(x2)<0 (x belongs to (0,1)).

f(x1)-f(x2)=1 x1+x1-1 x2-x2-x2 pass-score) = (x2+x2x1 2-x1-x1x2 2) x1x2(x2-x1)(1-xix2) x1x2

Because (x belongs to (0,1)).

So 1-x1x2 is greater than 0 and x2-x1 is less than 0

then f(x1)-f(x2)<0

Method 2: Derive the function.

f(x)‘=-x^(-2)+1

When f(x)'=0, x=1

Thus, on (0,1), f(x)' is less than 0, and on (1, positive infinity), f(x)' is greater than 0.

Thus f(x) decreases monotonically at (0,1).

Set (x1x2)-1>0.

f(x1)-f(x2)=1/x1+x1-1/x2-x2=(x2-x1)/(x1x2)-(x2-x1)=(x2-x1)[1/(x1x2)-1]>0。

So, f(x1) > f(x2).

i.e. f(x) is a subtractive function over the interval (0,1).

First, the definition method is used to make the difference, and the second is the derivative method.

The panacea for proving monotonicity: the derivative !! Take a derivative of this function and it will come out.

The derivative of f(x) is 1-1 (x 2), find the case where the derivative is greater than zero and less than zero.

That is, when 1-1 (x 2)>0, 1-1 (x 2)0, x takes (0,-1], [1,+infinity) is the increasing function.

When 1-1 (x 2).

f(x)=1+1/x

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

1/x1-1/x2

x2-x1)/(x1x2)

That is, f(x1)>f(x2), f(x) is drastically decreasing in the interval [1, + poor without mold holes].

Let 0 x1 x2, then.

f（x2）-f（x1）

1/x2+2）-（1/x1+2）

1/x2-1/x1

x1-x2）/（x1x2）

x1＜x2x1,x2＞0

f（x2）-f（x1）＜0

f（x2）＜f（x1）

f(x) decreases monotonically at x 0.

Summary. Proof that the function f(x)=-x-1) +2 is a monotonic subtraction function over the interval one to positive infinity.

On (0,+) f(x) is a number greater than zero, with f(x+1) f(x), its value is 1+1 x, because on (0,+, 1+1 x>1, so f(x+1) f(x), that is to say, f(x+1)>f(x), so it is monotonically increased.

(1) f(x) is monotonically decreasing on (0,1) and monotonically increasing on (1,+) proves: (1) either.

baidux1,x2 (0,1), and x10, so zhif(x) in (0,1) monotonically decreasing (2) any x1,x2 (1,+ and x11,1-1 x1x2>0 so f(x1)-f(x2) <0 so f(x) in (1,+ monotonically increasing (2) define the domain: x≠0, i.e. (- 0) (0,+ range: if you don't learn the basic inequality, you can use the discriminant method y=x+1 xx 2-yx+1=0 so δ=y 2-4 0 can solve the y -2 or y 2 value range (- 2] [2,+ play so hard, give more points.]

Figure 1 shows that at (1,+, either takes the two values of a and b, and a is shown in Figure 2 because 1 so a-1>0, b-1>0, b-a>0, so f(a)-f(b)>0, so the function f(x) is a monotonically decreasing function on (1,+).

Take x1 x2, and x1, x2 (1,+.)

f(x2)-f(x1)=1/(x2-1)-1/(x1-1)=(x1-x2)/(x2-1)(x1-1)

Since x1, x2 (1,+ so x2-1>0, x1-1>0, and x1-x2 0So f(x2)-f(x1) 0......That is, the function f(x) decreases monotonically over the interval (1,+