It is known that the numbers corresponding to two points a and B on the number axis are - 1 and 3 respectively, and point P is a moving point on the number axis, and its corresponding number is X 1) If the distances from point P to point a and point B are equal, find the number corresponding to point p; (point a is - 1, point O is 0, point B is 3, point P is the point between point 1 and point 2) (2) Whether there is a point P on the number axis, so that the sum of the distances from point P to point a and point B is only 6. If there is, we can find the value of X. if not, we can explain the reason

It is known that the numbers corresponding to two points a and B on the number axis are - 1 and 3 respectively, and point P is a moving point on the number axis, and its corresponding number is X 1) If the distances from point P to point a and point B are equal, find the number corresponding to point p; (point a is - 1, point O is 0, point B is 3, point P is the point between point 1 and point 2) (2) Whether there is a point P on the number axis, so that the sum of the distances from point P to point a and point B is only 6. If there is, we can find the value of X. if not, we can explain the reason

x1=-1,x2=3
（1）
xp-x1=x2-xp
xp=(x1+x2)/2=(-1+3)/2=1
（2）
|xp-x1|+|xp-x2|=6
|xp+1|+|xp-3|=6
When XP ＜ - 1: - xp-1-xp + 3 = 6, XP = - 2;
When - 1 < XP < 3: XP + 1-xp + 3 = 6, there is no solution;
When XP > 3: XP + 1 + xp-3 = 6, XP = 4
So XP = - 2, or 4
If AB = B-A = 4 and the distances from P to a and B are equal, then p is the midpoint of AB, P-A = B-P = 2, then p = 1
PA+PB=6，AB=4，
Point P is to the left of point a,
PA=A-P=-1-P，PB=B-P=3-P PA+PB=6 -1-P+3-P=6，P=-2
Point P is to the right of point B,
AP=P-A=P+1，BP=P-B=P-3 AP+BP=6 P+1+P-3=6，P=4。
（1）1
(2) Presence - 2 or 4
It is known that the numbers corresponding to two points a and B on the number axis are - 1 and 3 respectively, and point P is a moving point on the number axis, and its corresponding number is X. (2) whether there is a point P on the number axis,
Let the sum of the distances from point P to point a and point B be 6? If it exists, find out the value of X; if it does not exist, explain the reason
3) Point a and point B move to the right at two unit length / min and one unit length / min respectively, while point P moves to the left from point o at six unit length / min. when meeting point a, point P immediately moves to the right at the same speed, and constantly goes back and forth between point a and point B. when point a and point B coincide, what is the total distance of point P?
2. When x is a positive number
x-(-1) + x-3 = 6
x = 4
When x is negative
Similarly, x = - 2
So there are points P that are 4 and - 2, respectively
3. When AB overlaps, 4 / (2-1) = 4
No matter how P moves, the speed is constant, so the distance is 4 * 6 = 24 unit length
It is known that the numbers corresponding to two points a and B on the number axis are - 1 and 3 respectively. P is a moving point on the number axis, and its corresponding number is X
If the sum of the distances from point P to point a and point B is 6, then x =___
Just added Help to calculate thank
-2 or 4
The question was vague and didn't finish.
What about the picture
Let a belong to the R function f (x) = ax ^ 3-3x ^ 2, x = 2 is the extreme point of function y = f (x). Find the maximum value of function f (x) [- 1,5]
If f '(x) = 3ax ^ 2-6x, f' (2) = 12a-12 = 0, then a = 1, f (x) = x ^ 3-3x ^ 2
F '(x) = 3x ^ 2-6x = 3x (X-2), x = 0 and x = 2 are extreme points
f(-1)=-4、f(0)=0、f(2)=-4、f(5)=50.
Therefore, the minimum value of function f (x) in the interval [- 1,5] is - 4 and the maximum value is 50
The range of the function y = (cosx-3) / cosx + 3 is?
Let f (x) = - x ^ 3 + ax ^ 2 + (a ^ 2) * x + 1 (x belongs to R), where a belongs to R. when a is not equal to 0, find the maximum and minimum of F (x)
f'(x)=-3x^2+2ax+a^2=0
3x^2-2ax-a^2=0
(3x+a)(x-a)=0
X = - A / 3 or x = a
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When a
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y=loga(x^2+mx-m)(a>0,a≠1)
The true number is g (x) = x ^ 2 + mx-m
When the true number G (x) = x ^ 2 + mx-m gets all positive numbers, the range of y = loga (x ^ 2 + mx-m) is r
G (x) = x ^ 2 + mx-m opening upward
If the minimum value of G (x) = x ^ 2 + mx-m is greater than 0, for example, G (x) min = t > 0
Then G (x) = x ^ 2 + mx-m cannot take the positive number in the interval (0, t)
Only when G (x) min = t ≤ 0, G (x) = x ^ 2 + mx-m can get all positive numbers
The range is r
Then the true number gets all positive numbers
So the minimum value of true number is less than or equal to 0
Then the discriminant is greater than or equal to 0
So m ^ 2 + 4m > = 0
M=0
The domain is r
Then the true number is always greater than 0
So the discriminant is less than 0
m^2+4m
Given the function f (x) = 1 / 2x ^ 2-3x + 2lnx, it is proved that for any x1, X2 ∈ (0, + ∞), if X1 > X2, the inequality f (x1) - f (x2) > x2-x1 holds
When X1 > X2, f (x1) + X1 > F (x2) + x2g (x) = f (x) + X is to prove that G (x) is increasing when x > 0. G (x) = 1 / 2x ^ 2-2x + 2lnxg '(x) = X-2 + 2 / x > = 2 * radical (x * 2 / x) - 2 > = 2 radical 2-2 > 0 x > 0, so g (x) is increasing when x > 0