Judge the monotonicity of the function F X X 1 X over the interval negative infinity, 1 and prove

Solution: Increment function.

Let x exist on (- 1) any real numbers x1,x2 and x1y=f(x1)-f(x2).

x1/(x1+1) -x2/(x2+1)

x1+1)(x2+1)

x1x2+x1-x1x2-x2)/(x1+1)(x2+1)(x1-x2)/(x1+1)(x2+1)

x1-x2<0，(x1+1)<0,(x2+1)<0△y<0

So f(x)=(x+1) x is an increasing function over the interval (- 1).

Let -1 x2 x1, then x2-x1 0, x1x2 2, 1-1 x1x2 0

f(x2)-f(x1)=x2+1 x2-x1-1 x1=(x2-x1)+(1 x2-1 x1)=(x2-x1)(1-1 x1x2)>0, and the function f(x)=x+1 x increases monotonically in the interval (negative infinity, -1).

On (minus infinity, -1) take x1, x2, x2, > x1 so he x>0

y=f(x2)-f(x1)=x2+1 x2-x1+1 x11+1 x2-(1+1 x1)=1 x2-1 x1x1-x2 x1x2

Because x1x2>0, x1-x2>0

So he has to y>0

So f(x)=x+1 x increases monotonically over the interval (negative infinity, -1).

take x1, x2 belongs to this range, and x1 > x2

then f(x1)-f(x2).

x1-x2+1/x1-1/x2

x1-x2+(x2-x1)/x1x2

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

Because x1x2-1>0, x1-x2>0, x1x2>o, f(x1)-f(x2)>0

Therefore, f(x) is an increasing function in this interval.

Any x1 greater than x2 is at (negative infinity, -1), f(x1)-f(x2)=x1+1|x1-x2-1|x2=, and then pass the score, and that's it,

Summary. Formula deformation: x is a whole.

f(x)=1 x is an inverse proportional function.

Judge the monotonicity of the function f(x)=1 x over the interval (0, +infinity).

Hello, the function f(x)=1 x decreases monotonicity over the interval (0, +infinity).

Is there a process. This comes out at a glance, when x is bigger, y is smaller.

Okay, okay, how to judge whether f(x)=1 x is an even function?

Formula deformation: x is a whole.

f(x)=1 x is an inverse proportional function.

Summary. To prove monotonicity, we all prove it by the derivative function of the function, the interval where the derivative function is greater than zero, the original function increases monotonically, and the interval where the derivative function is less than zero decreases monotonically.

Prove the monotonicity of the function fx=x-+1 over the interval negative infinity to positive infinity.

Dear, what exactly is the function here?

This is the answer according to your question, do you see what you don't understand, if the topic is not this, please send the full question over.

To prove monotonicity, we all prove it by the derivative function of the function, the interval where the derivative function is greater than zero, the original function increases monotonically, and the derivative function is less than zero in the front, and the lead function of the original perturbation base decreases monotonically.

Dear, is there anything you don't understand here?

Summary. Hello dear, happy to answer your <>

To prove the monotonicity of the function $f(x)=x 2+1$ over the interval $(-infty, infty)$: it is necessary to prove its increase and decrease in the interval respectively. First, it is proved that $f(x)$ increases monotonically within $(-infty, infty)$.

Suppose there is $x 1, x 2$, where $x 1f(x 2)$, i.e., $(x 1) 2+1>(x 2) 2+1$. Move items to get $(x 1) 2>(x 2) 2$, which is the same as $x 1

Prove the monotonicity of the function fx=x-+1 over the interval negative infinity to positive infinity.

Hello dear, happy to answer your <>

To prove the monotonicity of the function $f(x)=x 2+1$ over the interval $(-infty, infty)$, it is necessary to prove its increase and decrease in the interval respectively. First, it is proved that $f(x)$ increases monotonically within $(-infty, infty)$.

Suppose Brother Cunqin is $x 1, x 2$, where $x 1f(x 2)$, i.e., $(x 1) 2+1>(x 2) 2+1$. Move items to get $(x 1) 2>(x 2) 2$, which is the same as $x 1

In mathematics, a function is a special relationship in which it maps each element in one set (called a "defined domain") to a unique element in another set (called a "value domain"). That is, a function can be thought of as a mapping that maps each element that defines a domain to a unique domain element.

Solution: f(x) is a subtractive function on [1,+.

Reason: Let x1 x2 1, then x1-x2 0, f(x1)-f(x2)=x1+1 x1-x2-1 x2

x1-x2)+(1/x1-1/x2)

x1-x2)+(x2-x1/x1x2)=(1-1/x1x2)(x1-x2)

x1-x2＞0

1-1/x1x2＜0

f（x）＜f（x2）

f(x) is a subtraction function on [1,+.

I don't understand, please ask, I wish you a happy o(o

What 1 x means is not very clear, the approximate method is as follows.

Use f(x+1)-f(x) to get a formula.

Then, on the interval where x belongs to [1, positive infinity), we discuss whether the formula is everstable in 0, and if everybody is in 0, then f(x+1)>f(x), that is, the original function is an increasing function in the interval.

If the constant is less than 0, the original function is a subtraction function over the interval.

Monotonically incremental. Let 1<=x1f(x1)-f(x2)=x1+1 x1-x2-1 x2=(x1-x2)(x1x2-1) x1x2<0

i.e. f(x1) so monotonically increasing.

Single bai decreasing on (0,1), du

In [1,+ on the single zhi to increase the tune, the following only proves the first dao one, the latter is basically the same as the inner front, let me not repeat it.

Take 00y=f(x2)-f(x1)=(x2+1 x2)-(x1-1 x1)=(x2-x1)-(1 x2-1 x1)=(x2-x1)-(x1-x2) (x1x2)=(x2-x1)(1-1 x1x2).

Because 00,1-1 x1x2<0, y<0, the function decrements on (0,1);

The derivative is followed by f(x).'=1+1 x2 , the derivative is greater than 0 so it is a monotonic function, and the combined image is a monotonic subtraction function.

Take x1, x2 in the f(x) field and 1 so (x1-x2) <010, x1*x2>1

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

That is, the function f(x)=x+1 x is an increasing function in the defined domain (1, positive infinity), and of course it can be solved by derivation.

is a monotonically increasing function.

f(x)=x+ 1 x (1, positive infinity) f(x+1)=x+1+ 1 (x +1)(1, positive infinity) f(x+1)-f(x)=x+1+ 1 (x +1)-(x+ 1 x).

1- 1/[x（x+1)]

Because of x1, x(x+1) 2

So 1 [x(x+1)] 1

So 1- 1 [x(x+1)] 0

Therefore, f(x) is a monotonically increasing function.

Increasing!!!

Because: Method 1: Derivative f'(x) = 1-1 (square of x) because 1 (square of x) is in (1, positive infinity.

Large) on less than 1, so f'(x)=1-1 (the square of x) is greater than 0 on (1, positive infinity), i.e., f(x)=x+one part of x is increasing on (1, positive infinity).

Method 2: It can be proved by a combination of inequality and drawing that it is a tick function.

Derivative f(x)."=1-1/x^2

Because x>1

So f(x).">0

So f(x)=x+x/1 increments over (1, positive infinity).

f(x)'=1-1 x 2 allows f(x)>0 to solve the equation.

Didn't you have the original question? You can do the derivation! See if the derivative is greater than zero or less than zero in the interval! Greater than zero is an increase range! Less than the zero minus range!

Monotonically increasing, f(x) derivatives 1-1 x2 at (1,+ evergrande, at 0

Find the derivative of f(x) and balance greater than 0 from 1 to positive infinity, so it is single.

Prove by definition: Let x1, x2, x1 be less than x2, and f(x2)—f(x1) is greater than 0, then it can be proved.

The function is single-incremented, x=1 is the minimum value, and x=4 is the maximum value.

It is directly proved by the definition method, monotonically increasing.

Since at (1, positive infinity) is an increasing function, f(1)min and f(4)max

is f(x) = (x 2+1)-x

If this is the case, then at (- 0) is monotonically decreasing f(x) = (x 2+1)-x=1 (x 2+1)+x because at x (-0), (x 2+1)> x 2>|x|, so (x 2+1)+x>0, when x1>x2, and x1, x2 (-0), [ x1 2+1)+x1]-[x2 2+1)+x2]=(x1-x2)[

x1+x2)/√（x1^2+1）+√x2^2+1)-1]