It is proved that y = x - [x] is a periodic function, and the minimum positive period is?

It is proved that y = x - [x] is a periodic function, and the minimum positive period is?

Where are the steps? This problem (higher non basic function) can only be solved by graph. It is a line segment with a slope of 1 in each grid. The period also comes from this. The minimum positive period is 1

What is the relationship between the existence of the limit of the function at point X and the continuity of the function at point X and the uniform continuity of the function at point x?

The existence of the limit of the function at point x means that the left and right limits of the point exist and are equal, regardless of whether the point is defined or not and the value of the function
If the function is continuous at point x, it means that the limit of this point exists and is equal to the function value of this point
In general, it seems that only functions have uniform continuity in an interval, not at a point

When x0, f (x) = xsinx / 1, the continuous interval of this function is

Just consider x = 0
X approaches zero from the left, and the x ^ (- infinity) limit is 0; X approaches zero from the right, and the number multiplied by the absolute value of no more than one is zero; When x = 0, f (x) = 0, so the continuous interval is [negative infinity, positive infinity]

The monotonically decreasing interval of function f (x) = e ^ xsinx in (- π, π) F(x)=e^xsinx F’(x)= e^xsinx+e^xcosx =e^x(sinx+cosx) = e^x√2（√2/2*sinx+√2/2*cosx） =e^x√2(sinπ/4sinx+cosπ/4cosx) =e^x√2cos(x-π/4) Just discuss its sign. What is it? Please explain in detail

When x ∈ (- π, - 3 / 4 π], f '(x) < 0, f (x) decreases monotonically
When x ∈ (- - 3 / 4 π, 3 / 4 π), f '(x) > 0, f (x) increases monotonically
When x ∈ (3 / 4 π, π), f '(x) ＜ 0, f (x) decreases monotonically

The monotonically decreasing interval of function f (x) = (x-3) ex is () A. （-∞，2） B. （0，3） C. （1，4） D. （2，+∞）

∵ number f (x) = (x-3) ex
∴f′（x）=（x-2）ex，
According to the relationship between monotonicity and inequality:
(X-2) ex < 0, i.e. x < 2
So the monotone decreasing interval of function f (x) = (x-3) ex is (- ∞, 2)
Therefore: a

Steps for judging function parity

The first step of judging the function: find 1. If the definition field is about, find f (- x) to see its relationship with F (x). 2. If the definition field is asymmetric about the origin, it can be said that the function is the second step: see f (- x) its relationship with F (x). If f (- x) = - f (x), the function is. If f (- x) = f (x), the function is. Note: find the purpose of the definition field. 1. See