What is the monotonic reduction interval of f x x 1 x, x 0 ? Let s talk about the steps

Hehe, this is a classic question, a must-have brain before the college entrance examination

I have learned to seek guidance before the college entrance examination, so it is beneficial to get in touch with it a little ahead.

Since a function is a subtraction function in the interval where the derivative is less than 0, it is good to find the derivative of the function.

Abbreviated as "derivation"].

Solution: Derivative f(x)=x+1 x,(x 0) on both sides at the same time to obtain.

f'(x) = 1 - (1 x 2) because f'(x) < 0, 1 x 2>1, and x 2<1, that is, 0<=x<1, and because x>0, so 0 In summary, the monotonic reduction interval of f(x)=x+1 x(x 0) is (0,1).

This is a math problem in the first year of high school, and I forgot what a single interval is.

Upstairs there's a derivative, advanced mathematics.

If you want to kill him, why don't you come out with the whole calculus.

This is called a tick function.

Find the two vertices (1,2) (-1,-2).

On (0,1) is a subtraction function.

I don't have the conditions to draw an image, sorry.

Repair! How to find the vertex!

Let x=1 x, find two x, bring in, find two y's are vertices.

In this comparison of typical problems, you need to know the steps of doing a monotonous problem of functions, one is to define the domain, the second is parity, and the third is to do it; My steps are as follows: The definition is that x is not equal to 0, and it is an odd function, so only consider the case where x is greater than 0, and the following can be done in a variety of ways, one is the definition method, and the other is to find the derivative.

Let's do it by yourself below.

Provide the mind to learn by yourself

Huh

Order f'(x)=x,f"(x) = 1 x, the intersection of the two is (1,1), so the monotonic reduction interval is [0,1].

Derivation. f'（x）＝3x²-3

Reel 3x -3 0

Solve -1 x 1

Therefore, the monotonic reduction interval is [-1,+1].

f(x)=log3(1-x)+[1 (x-1)] The domain of the function log3(1-x) is: (-1) function y=log3(1-x) can be split into: y=log3(t) (monotonically increasing) t=1-x (monotonically decreasing) so the first half of the function log3(1-x) in f(x) is a single-plex digging modulation function, and the second half of the function infiltrating ridge nucleus 1 (x-1) is the third field nucleus centered at point (1,0).

When x 0, f(x)=x 2-2x-1=(x-1) 2-2 image is a parabolic arc on the right side of the y-axis with x=1 as the axis of symmetry and the opening facing upwards f(-x)=x 2-|-x|-1=f(x), f(x) is an even function, the image is symmetrical about the y-axis, and the image on the right side of the y-axis can be combined with the image along the y-axis, and the monotonically decreasing interval of f(x) is obtained.

y=1/3x³+x²-8x

y'=x +2x-8=(x+4)(x-2)let y'=0, x=2 or x=-4

When x<-4 or x>2, y'>0 and y is the increment function.

When -4 therefore the increase interval for y is (-4) and (2,+ the decrease interval is [-4,2].

y=1/(1-x)

From the denominator is not 0,1-x≠0,x≠1, the domain of the function is defined as.

When x (1), as x increases, 1-x decreases, 1 (1-x) increases, and (1) is the increasing interval of the function y;

In the same way, (1, is also the increasing interval of the function y, therefore, the function y has no decreasing interval.

The domain of the f(x)=log3(1-x)+[1 (x-1)] function log3(1-x) is as follows: (-1) function y=log3(1-x) can be split into the following

y=log3(t) (monotonically incremented).

t=1-x (monotonically minus).

So the first half of the perturbation function log3(1-x) in f(x) is a monotonic subtraction function, and the second half function 1 (x-1) is a hyperbola centered in the third quadrant at point (1,0), which is also a subtraction function, so.

The original function is subtractive over the entire defined domain;

The monotonous reduction zone blocking judgment is as follows: (-1) Jane Li Gai.

Answer: f(x)=|x²-1|+x

1) When x -1 < = 0 i.e. -1 < = x < = 1:

f(x)=1-x +x=-(x-1 2) +5 4 The monotonically decreasing interval is [1 2,1].

2) When x -1 > = 0, i.e. x < = -1 or x > = 1:

f(x)=x -1+x=(x+1 2) -5 4 The monotonically decreasing interval is (- 1].

In summary, f(x)=|x²-1|The monotonically decreasing interval of +x is [1 2,1] or (- 1).