Let f be a monotone function on R, define g (x) = f (x + 0), and prove that the function g is right continuous at every point on R ∵ f is a monotone function on R, and the discontinuous point of ∵ f is the most countable set I'm a freshman. I haven't mentioned this conclusion in my textbook so far. How can I prove it

prove:
∵ f is a monotone function on R,
The discontinuous points of F are the most countable set
Let the set of discontinuous points of f be e
Then G (x) = f (x + 0) = f (x) x ∈ R / E
Therefore, if G (x) is continuous on R / E, it is also right continuous
Let x ∈ e, G (x) = LIM (T - > x +) f (T) = LIM (T - > x +) g (T) t ∈ R / E
х G is right continuous in X
In conclusion, G (x) is right continuous over R
Supplement: the proof does not use whether there is such an open interval. However, it can be explained as follows
Because e is contained in R, and E is the most countable set
Then R-E is thicker than e, that is, every point in E can be approximated by the point sequence in r-e
Since LIM (T - > x +) f (T) exists, the limit obtained by let t belong to R / e must be the same as that obtained by t ∈ R. therefore, it is proved as above
Note that D (x) is not monotonous
On that step, we use the conclusion of real variable function
Supplement3: I will not^^

It is proved that FX = x / X-1 is a monotone decreasing function on (1, positive infinity) by the definition of function monotonicity

Certification:
f(x)=x/(x-1)=1+1/(x-1)
In (1, + ∞), take x1, x2
Let 1F (x2)
So the function f (x) = x / (x-1) is a monotone decreasing function on (1, positive infinity)

The function f (x) = x + 1 / X is known and belongs to [2,5]. The monotonicity definition is used to prove that FX is increasing in the interval [2,5]

Let X1 and X2 be arbitrary values in the function interval [2,5], then let X1 < X2, then:
f(x1)-f(x2)=x1/(x1+1)-x2/(x2+1)=(x1-x2)/(x1+1)X(x2+1)
Because 2 ≤ X1 < x2 ≤ 5
So x1-x2 < 0, X1 + 1 > 0, X2 + 1 > 0
So f (x1) - f (x2)

How to do the checking problem of 60 divided by 7

60/7=8…… four
Checking calculation: 7x8 + 4 = 60

Given that the inclination angle of the straight line m is twice of that of the straight line root 3-3y-3 = 0, and the intercept of the straight line m on the X axis is - 3, then the equation of the straight line m is derived

Should there be an "X" after the first "3"? K1 = 1, θ 1 = 45 & # 186;; θ 2 = 90 & # 186; m slope does not exist, the equation of the straight line M: x = - 3 is obtained. Or "3" is the error of "X"? K1 = 1 / 3, K2 = 2K1 / (1-k1 & # 178;) = (2 / 3) / (1-1 / 9) = 3 / 4 from the point oblique equation: y-0 = K2 (x-a) = >

(2.5x + 20) divided by (x + 20) = 2 to find the solution of the equation (x is unknown)

(2.5x + 20) divided by (x + 20) = 2
2.5x/x+20=2/1
2.5x + 20 = 2 times (x + 20) (cross multiplication)
2x+40=2.5x+20
0.5x=20
x=40

Given the set a = {x │ x ^ 2 + (M + 2) x + 1 = 0}, and a ∩ R + = empty set, find the value range of real number M
Here's what I do:
Because y = x ^ 2 + (M + 2) x + 1 is constant crossing point (0,1), so ⊿ > = 0, so m = 0
And because the abscissa of vertex (M + 2) / (- 2) - 2
In conclusion, the range of M is (- 2, - 4] ∪ [0, + ∞)
But the answer is (- 4, + ∞)
What's wrong with me?

Here the set a = {x │ x ^ 2 + (M + 2) x + 1 = 0} a ∩ R + = empty set, which means:
The function y = x ^ 2 + (M + 2) x + 1 has no intersection with the positive semiaxis of X axis
1. There is no intersection between Y and X axis, ⊿ - 4

Simple calculation of 31.1 / (3.11 * 2.5)

31.1/(3.11*2.5)
=3.11*10/(3.11*2.5)
=10/2.5
=100/25
=4
Do not understand welcome to ask

If the point a (- 3,2m-1) is symmetrical about the origin in the fourth quadrant, then the value range of M is_________ ?

M>1/2

One half plus two thirds plus three quarters plus four fifths plus five sixths plus six sevens

One half plus two thirds plus three quarters plus four fifths plus five sixths plus six sevens =(1/2+2/3+3/4+4/5+5/6+6/7)²=(6/12+8/12+9/12+10/12+28/35+30/35)²=(11/6+58/35)²=(525+348/210)²=(291/70)²=...