Find the monotonicity of y x 1 x

First of all, x is not 0, and the function is divided into two parts, first: y=x, second: y=1 x", so that it is easy to analyze the monotonicity of the division function.

Just analyze the second function. The monotonicity of y=1 x is: monotonically decreasing in the left semi-axis increases monotonically in the semi-axis, and is meaningless at the zero point, so the function increases monotonically in the left and right semi-axis.

Satisfied?

Set x1>x2>0

So f(x1)-f(x2)=x1+1 x1-(x2+1 x2)x1-x2+(x2-x1 x1x2).

x1-x2)(1-1/x1x2)

x1-x2)(x1x2-1/x1x2)

Because x1x2>0 x1-x2>0

1.When x1x2 > 1, x1x2-1>0

So x1>x2>1, so when x>1 the function is incremental.

2.When x1x2 > 1, x1x2-1>0

So 0 minus function.

So when x>1 the function is incremental, so when 0When < x<1, the function is subtractive.

Find the first derivative y' = 1 - 1/(x^2)y'> = 0 monotonically incrementing, y'<=0 is monotonically decreasing.

If you don't learn derivatives, you can only judge by definitions.

f(x1)-f(x2) = ..

Monotonous increase, the method is as follows, please refer to:

Let's do the derivative, and then see whether the derivative is greater than zero or less than zero, so that you can find the monotonicity of the function, but this function x is not equal to zero, so I think the discussion should be disconnected at the point of 0.

My method is to look at them one by one: x is a proportional function, monotonically increasing, one-tenth of x is monotonically decreasing, adding a negative sign is monotonically increasing, and increasing increases to increase, so y is an increasing function. (Optional).

Hello. You can use the definition method, or you can directly use the first derivative to find the derivative, after the derivative is 1+x2, it must be a number greater than zero is monotonically increasing

y = x²-4x+1 = x-2)²-3

The decreasing interval of the age of the letter is (- 2), and the increasing range is (2, rent + ).

Analysis The original function is a composite function of the inner function t=x2-4x+1 and the outer function g(t)=(12)t(12)t, the monotonic interval of the inner function is obtained, and then the monotonicity of the composite function is used to obtain the monotonic interval of the original function, and then the range of t is obtained by using the collocation method, and the value range of the original function can be obtained by substituting the outer function

Answer: Let t=x2-4x+1, then the original function is g(t)=(12)t(12)t, the decreasing interval (- 2] of the number of inner Han Zen buried auspicious ridges t=x2-4x+1, the increasing interval is (2,+), and the outer function g(t)=(12)t(12)t is the subtraction function, the increasing interval of the original composite function is (- 2], and the decreasing interval is (2,+).

t=x2-4x+1=(x-2)2-3 -3,g(t)=(12)t(12)t (0,(12) 3(12) 3]=(0,8].

The function y=(1212)x2-4x+1 is in the range of (0,8].

For discussing monotonicity, using derivative is the most macro and fastest.

The derivative y, =1-1 x 2, so that y, suffocating beam = 0, then x=1 or -1 When x belongs to (negative infinity, -1) (1, positive infinity), y, > 0, i.e., y increases monotonically.

When x belongs to (Antine-1,1), y,

The domain in which the function is defined is x r

Let x10f(x1) -f(x2)>0, i.e., f(x1)>f(x2) f(x) monotonically decreasing on (- 0).

In the same way, let 00 (x1 + x2) (x1-x2) < 0

f(x1) -f(x2)<0, i.e.: f(x1).< f(x2) f(x) is monotonically increasing on (0,+.

The domain of the function is (- take two points x1, x2 on (- 0), and x1o f(x2)-f(x1).

f(x2) so the function is monotonically reduced at (- 0).

In the same way, the provable function increases monotonically at (0,+.

Solution: Definition method to solve the problem.

y=x²-1

Let x1 have f(x1)-f(x2).

x1²-1）-（x2²-1）

x1²-x2²

x1-x2）（x1+x2）

It can be obtained from the title.

x1-x2<0

When 0 we have x1+x2 0

There are f(x1)-f(x2)<0

i.e. f(x1) increases monotonically.

The same can be said. When x1 has f(x1) f(x2).

Monotonically decreasing.

When x 0, it decreases monotonically.

When x 0, it is monotonically incremented.

This can be drawn or defined as monotonous.

If you have learned derivatives, you can use the method of derivation.

This function means that the image squared by y=x is shifted one unit downwards on **, so it is still a quadratic function, roughly presenting a V-shape, with the y-axis as the axis of symmetry, so, on the left side of the y-axis, it is monotonically decreasing, and on the right-hand side of the y-axis, it is monotonically increasing.

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:

Monotonic increase interval: (-1,0),(a -2a,+infinity)Monotonic decrease interval:(0,a -2a).

When a = 0 or 2, f (x) = x 0 monotonically increases the interval (-1, +infinity).

When 0 a1 or 1 a2, 0 a(2-a) 1, at this time:

Monotonic increase interval: (-1,a -2a), (0,+infinity) Monotonic decreasing interval: (a -2a,0).

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

Monotonic increment interval: (0, +infinity).

y=x(x^2-1)=x³-x

Derivation. y'=3x²-1

Cause. y'=3x²-1＜0

Find the monotonic decreasing interval of the epoch function as.

3/3＜x＜√

y'=3x²-1＞0

The monotonic increase interval of the function is . x＜

3 3 or x

So the monotonicity of the function in the interval [0,1] is:

The monotonic envy segment minus interval is. 0＜x＜√

The monotonous increase in the reputation interval is.

3/3＜x＜1

The meaning of this function is that the image squared by y=x is shifted one unit downwards on **, so it is still a quadratic function to do bands.

It is roughly V-shaped, with the y-axis as the axis of symmetry.

So, on the left side of the y-axis, it's monotonically decreasing, and on the right-hand side of the y-axis, it's monotonically increasing.

known: y=1 (x 2x 3);

then y'=(2

2x) [x 2x 3)] 2 (the derivative formula for division should be fine);

Lingy'=0 gets: x=1;

When x0 is at this time, the original function is single, and the attack increases;

When x>1, y'As for x=1, you can feel free to put it in x>1 or x if you haven't learned derivatives yet:

y=1 (x 2x 3)=1 [(x-1) 2+2] At this point, what we want to investigate is the increase and decrease of (x-1) 2, when x is the same as the difficult brother, when the vertical x>1, the original function is single-decreased.

y=[(1/2)^x]^2-2*(1/2)^x+2;

(1/2)^x-1]^2+1;

known by the monotonicity of quadratic functions;

1 2) (x<0) monotonically increasing at x>1; and (1 2) x is decreasing; So x<0 decreases monotonically;

1 2) (x>0) monotonically decreasing at x<1; and (1 2) x is decreasing; So x>0 increases monotonically;

It can be solved with derivatives.

The derivative of the original function is y'=1-1/[2√(x-1)],x>1。Let the derivative be greater than or equal to 0, i.e., monotonically increasing, and solve x>=5 4, so that the function is an increasing function on [5 4, positive infinity), and the same can be seen that the function is a subtraction function on (1, 5 4).

Solution: Let f(x)=y=x (1+x).

f'(x)=[x'(1+x²)-x(1+x²)']/(1+x²)²=(1-x²)/(1+x²)²

Order f'(x) 0, get (1-x) (1+x) 0x -1 0

The monotonically increasing interval of the 1 x 1 function is [-1,1].

The monotonically decreasing interval of the function is (- 1], [1,+