Find the monotone interval of the following function (1)y=x-Inx (2)y=1/2x

Find the monotone interval of the following function (1)y=x-Inx (2)y=1/2x

1) The domain is defined as x > 0
Y '= 1-1 / x = (x-1) / x, the extreme point is x = 1
Monotone increasing interval: x > 1
Monotone decreasing interval: (0,1)
2) If y = (1 / 2) x, then it is a positive proportional function, monotonically increasing on R;
If y = 1 / (2x), then it is an inverse proportional function, which is monotone decreasing on (- ∞, 0), (0, + ∞)

Find function monotone interval f(x)=|x^2-3x+2|

f(x)=|(x-3/2)²+2-9/4|=|(x-3/2)²-1/4|=|(x-1)(x-2)|
(- ∞, 1] minus
(1,3 / 2] increased
(3 / 2,2] minus
(2, + ∞) increase

Are monotone functions bounded on closed intervals?

Closed interval continuous function must be bounded and monotone function bounded

It is proved that a monotone function on a closed interval is bounded, and that a monotone function on an open interval is not necessarily bounded

Let f (x) monotonically increase on the interval [a, b], if x ∈ [a, b] f (a) < = f (x) < = f (b), then f (x) is bounded
Example f (x) = 1 / X is unbounded in the open interval (0,1)

Is monotone function bounded? Is continuous function bounded?

(1) Monotone functions are not necessarily bounded
For example, the exponential function f (x) = e ^ x is monotonically increasing in its definition domain (- ∞, + ∞),
But obviously it has no upper bound, thus unbounded!
(2) Continuous functions are not necessarily bounded
For example, also consider the exponential function f (x) = e ^ x, (- ∞, + ∞), which is a basic elementary function,
So it must be continuous, but obviously it's unbounded!

How to prove the boundedness of function? For example, it is proved that y = xcosx is unbounded in the range of real numbers. I only know that x = 2K Π, K can take infinity, then the value of function is infinite. How to write it?

If the function is bounded, there is a contradiction between | y | m for any x, so the function y = xcosx is unbounded

1. Is a monotone function necessarily continuous on a closed interval? 2. If a function can take all the points on a closed interval, is it bounded?

1: Not necessarily. If x is not equal to a number in this closed interval, there is a breakpoint
2: It is not when it is a piecewise function, such as: F (x) = 1 △ x, 0 ＜ x ＜ = 1; f (x) = 1 △ 2, x = 0

If four functions are given, they have the following two properties: ① the minimum positive period is π; ② the image is about point (π) The symmetric function is () A. y=cos（2x-π 6） B. y=sin（2x+π 6） C. y=sin（x 2+π 6） D. y=tan（x+π 3）

The minimum positive period of the function is π, so π = 2 π
|ω|, according to the option, ω > 0, so ω = 2, excluding C
Image about point (π)
So x = π
At 6, the function value is 0
Obviously, a and B don't satisfy the question, π
6 +π
3=π
Two
y=tan（x+π
3) The center of symmetry of is (π
6，0）
Therefore, D is selected

On two function images of symmetry of origin Is there a relationship like odd function That is, if FX is equal to negative g minus X

There is y = f (x) equal to y = - f (- x)
Definition: for a function, it is symmetric about the origin (0,0) in the definition domain, and satisfies 1 for any X. in the odd function f (x), the signs of F (x) and f (- x) are opposite and the absolute values are equal, that is, f (- x) = - f (x), on the contrary, the function y = f (x) satisfying f (- x) = - f (x) must be an odd function. For example, f (x) = x ^ (2n-1), n ∈ Z; (f (x) is equal to the 2N-1 power of X, n belongs to an integer)
2. The image of odd function is symmetric about the center of origin (0,0)
3. 4. If f (x) is an odd function and X belongs to R, then f (0) = 0
5. Let f (x) be differentiable on I. if f (x) is odd function on I, then f '(x) is even function on I. that is, f (x) = - f (- x) can be derived from F' (x) = [- f (- x)] '(- x)' = - f '(- x) (- 1) = f' (- x)

What is the symmetry between the two images of the function y = x ^ 3 and y = x ^ (1 / 3),

It's symmetric about y = x, because y = x ^ (1 / 3) if you take y as an independent variable, it's actually x = y ^ 3