f x x3 1 proves monotonicity

Let x1 and x2 be any two points in the defined field, and x10, when x1 and x2 are the same sign, f(x1)-f(x2)>0

When x1 and x2 are different signs, x2 2+x1 2>=2x1x2, so x2 2+x1 2+x1x2>=3x1x2>0, f(x1)-f(x2)>0

F(x1)-F(x2)>0

So it's a subtraction function.

This one is very simple, proving that monotonicity is a routine.

Let any two numbers x1 and x2, where x1 is greater than x2, and you only need to prove that f(x1) is less than f(x2).

That's the end of it.

If I say that you don't know how to do this, you don't want to study math anymore, it's a waste of time!

Let x1>x2,f(x1)-f(x2)=-x1 3+x2 3=(x2-x1)(x1 2+x1x2+x2 2)=(x2-x1).

x1>x2, x2-x1<0, and [x1+(x2 2)] 2+3x2 2 4>0

So f(x1)-f(x2)<0, so f(x) is a subtraction function.

Take two points a and b on the function, and a > b

f(a)=-a^3+1

f(b)=-b^3+1

f(a)-f(b)=b^3-a^3

Because a>b, f(a)-f(b)=b 3-a 3<0 gives f(x)=-x3+1 is a subtraction.

Let x2 6 9x1 f(x2)-f(x1)=-3(x2)+3(x1)=3(x1-x2) 6 80 So the function decreases monotonically.

x1,x2∈(-x1>x2

f(x1)-f(x2)

x1^3-x2^3

x1-x2)(x1^2+x1x2+x2^2)(x1-x2)[(x1+x2/2)^2+(3/4)*x2^2]x1>x2 x1-x2>0

x1+x2 2) 2+(3 4)*x2 2>0 shows that f(x) increases monotonically on r.

f（x）=3x-x^3

f'Wheel width (x) = 3-3x 2

Order f'Iwaming (x) = 0

3-3x^2=0

x= 1 when x0 monotonously increases the brightness of jujubes.

When x>1 f'(x)

Of course, it is more troublesome to use the definition proof, before the definition proof, first introduce two methods method one: y=x 3 y=x are monotonically increased, which can be clearly seen on the function image, then f(x)=x 3+x monotonically increases, which is suitable for quickly solving multiple-choice questions, or fill-in-the-blank and other problems that do not require a process.

Method 2: Derivative method. f(x)=x 3+x, so f'(x)=3x 2+1 is obviously greater than 0, so it increases monotonically on the entire x-axis, which is simple and straightforward.

Below, we will use the definition to prove it:

Take x1 then: f(x1)-f(x2)=x1 3-x2 3+x1-x2=(x1-x2)(x1 2+x1x2+x2 2)+(x1-x2).

x1-x2) <0

So, x1When < x2, f(x1)-f(x2)<0, then the function increases monotonically.

Solution: Set x1 x2

f(x1)-f(x2)=(x1)³+x1 - x2)³-x2=(x1)³-x2)³+x1-x2

x1＜x2(x1)³-x2)³＜0，x1-x2＜0∴f(x1)-f(x2)＜0

That is, f(x) is a monotonic increasing function.

Reel x1>x2

f(x1)-f(x2)=(x1 3-x2 3)+(x1-x2)=(x1-x2)(x1 2+x1x2+x2 2)+(x1-x2)=(x1-x2)(x1 2+x1x2+x2 2+1)=(x1-x2)[(x1+x2 2) 2+3x 2 4+1]x1>x2, so x1-x2>0

x1+x2 2) 2+3x 2 4+1, two squares plus 1, so greater than 0

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

That is, f(x1) > f(x2) when x1 > x2

So it's an increment function.

f(x1)-f(x2)=x1^3-x2^3=(x1-x2)(x1^2+x1x2+x2^2)

Cubic Variance Formula).

Because x1 cores let 0

x1 and x2 belong to the set of real numbers.

The width of Sola's is f(x1)-f(x2)<0 for any real number x1, i.e. f(x)=x 3 in. r

on monotonically incrementing.

Prove that x1 and x2 belong to r, and x1 x2

then f(x1)-f(x2).

x1^3-x2^3

x1-x2)(x1^2+x1x2+x2^2)=(x1-x2)[x1^2+x2x1+(x2/2)^2+3x2^2/4]

x1-x2)[(x1+x2 2) 2+3x2 2 4] from x1 x2 to x1-x2 0

If x1 and x2 cannot be 0 at the same time, then (x1+x2 2) 2+3x2 2 4 0

then (x1-x2)[(x1+x2 2) 2+3x2 2 4] 0 then f(x1)-f(x2) 0

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

Then f(x)=x 3 is an increase function on r.

Certificate: Let x1, x2 r, and x1 x2, then:

f(x)?f(x

x?x(x?x)(x

x+xx)=12(x

x)[(x+x)

x+x；x1 x2, x1-x2 0, x1, x2 are not all 0, (x+x)+x+x

0；∴f（x1）＜f（x2）；

f(x)=x3 is an increment function over r

Let x1,x2 [1,2], and 10,00,(1-4 x1x2)>0 (x2-x1)(1-4 x1x2)>0

(x2-x1)(1-4 x1x2)<0, i.e., f(x1)-f(x2)<0

Monotonically decreasing on 1,2.

f(x) = ln(x+ 1) -ax (x+a) = ln(x+ 1) -ax+a2-a2) (x+a) = ln(x+ 1) -a + a2 (x+a) define the domain: x -1, and x≠-a f (x) = 1 (x+1) -a2 (x+a)2 = = = x When a 0 or a 2, (2-a) 0, at this time: monotonically increasing interval:

1,0),(a2-2a,+infinity) monotonically decreasing interval:(0,a2-2a) when a=0 or 2, f (x) = x2 0 monotonically increasing interval (-1,+infinity) When 0 a1 or 1 a2, 0 a(2-a) 1, at this time: monotonically increasing interval:

1,a2-2a),(0,+infinity)monotonic reduction interval:(a2-2a,0) When a=1:f (x) = x (x+a)2 0 monotonic reduction interval:

1,0) Monotonic increase interval: (0,+infinity).

(- 0) and (0, ) are increments, respectively.

f'=3*x 2+1 x 2>0 holds on (- 0) and (0, ), respectively, so on (- 0) and (0, ), respectively, are increment functions.

It is worth noting that it cannot be written.

0)u(0, ) is an increasing function, because f is not necessarily larger on (0, ) than on (- 0), so two intervals cannot be regarded as a whole (- 0) u(0, ) as an increasing interval.

By definition, pay attention to the case-by-case discussion.

1 When x is less than 0, so that x1 is less than x2, then f(x1)-f(x2)=Sort it out yourself.

Get the equation for x1 and x2, and then use the condition that x1 is less than x2 to see if it increases or decreases.

2. When x is greater than 0.

Just tell you the general steps and calculate it yourself, that's how to do it. Hope!