Some rational numbers cannot be expressed on the number axis

Some rational numbers cannot be expressed on the number axis

Error, all real numbers can be expressed on the number axis, whether rational or irrational!
Solving rational numbers can be represented by the points on the number axis. What kind of mathematical thought does it embody?
There is also a thank you answer; two numbers that are opposite to each other are related to the corresponding points on the axis____ Symmetry. They're here_____ The distance is equal
It embodies the mathematical thought of the combination of number and shape,
The corresponding points of two opposite numbers on the number axis are symmetrical about the origin. They are equidistant from the origin
3 times the root number 3 times the root number 2
3 times root number 3 times root number 2
√(3√(3√2))
=(3√(3√2))^(1/2)
=3^(1/2)*(√(3√2))^(1/2)
=3^(1/2)*(3√2)^(1/4)
=3^(1/2)*3^(1/4)(√2)^(1/4)
=3^(3/4)2^(1/8)
The function f (x) = AX3 + bx2 + CX obtains the minimum value 5 at the point x0, and the image of its derivative goes through (1,0), (2,0), as shown in the figure, to find: (1) the value of x0; (2) the value of a, B, C; (3) the maximum value of F (x)
(1) It can be seen from the image that f '(x) > 0 on (- ∞, 1) and f' (x) < 0 on (1, 2). F '(x) > 0 on (2, + ∞). Therefore, f (x) increases on (- ∞, 1), (2, + ∞) and decreases on (1, 2). Therefore, f (x) gets a minimum at x = 2, so x0 = 2. (2) f' (x
Is 2 / 2 of the root a fraction?
no
Because 2 / 2 of the root equals 2
The minimum values of the function (x 2, x 0, x 2) = (x 2, x 0, x 2) are obtained at the point of (x 2, x 2), x 2 + x 1
(1) It can be seen from the image that f '(x) > 0 on (- ∞, 1) and f' (x) < 0 on (1, 2). F '(x) > 0 on (2, + ∞). Therefore, f (x) increases on (- ∞, 1), (2, + ∞) and decreases on (1, 2). Therefore, f (x) has a minimum value at x = 2, so x0 = 2. (2) f' (x) = 3ax2 + 2bx + C. from F '(1) = 0, f' (2) = 0, f (2) = 5, 3A + 2b is obtained +C = 012a + 4B + C = 08A + 4B + 2C = 5, the solution is a = 52, B = - 454, C = 15; (3) from (1) we know the maximum value of the function at x = 1, f (1) = 254
Is the root two of two a fraction
no
Because the root 2 is an irrational number and the rational number includes fraction, it means that fraction is only in the category of rational number
So the root of two is not a fraction, but the simplest quadratic radical rationalized by denominator
Given that the image of the function f (x) = ax ^ 3 + BX ^ 2 + CX + D (a ≠ 0) passes through the origin, f '(1) = 0, if f (x) reaches the maximum value 2 at x = - 1
(1) Find the analytic expression of the function y = f (x); (2) if f (x) ≥ f '(x) + 6x + m for any x belonging to [- 2,4], find the maximum value of M
F (0) = 0, d = 0, f (x) = ax ^ 3 + BX ^ 2 + CX = ax (x ^ 2 + BX / A + C / a) because the function has the maximum value when x = - 1, it has the following properties: F (0) = 0, d = 0, f (x) = ax ^ 3 + BX ^ 2 + CX = ax (x ^ 2 + BX / A + C / a)
(b/a)^2-4c/a=0,b^2=4ac (1)
And f (- 1) = 2, - A + B-C = 2 (2)
f'(x)=3ax^2+2bx+c,f'(-1)=0,3a-2b+c=0 (3)
Solutions (1), (2), (3) are obtained
a=-1/2,b=-3,c=-9/2.
f(x)=-x^3/2-3x^2-9x/2
f'(x)≥f'(x)+6x+m,6x+m≤0,x≤-m/6,m≤-6x,
And - 2 ≤ x ≤ 4, m ≤ - 24. M, the maximum value is - 24
If f (0) = 0, then d = 0;
f'(x)=3ax^2+2bx+c
If f '(1) = 0, then 3A + 2B + C = 0, ②;
If f (x) has a maximum of 2 at x = - 1, then f '(- 1) = 0, 3a-2b + C = 0, ③;
And f (- 1) = 2, - A + B-C = 2, 4;
From the above four formulas, it can be concluded that:
a=1，b=0，c=-3，d=0
1: Through the origin, d = 0
f'(x)=3ax^2+2bx+c；f'(1)=3a+2b+c=0
If f (x) has a maximum of 2 at x = - 1, then f '(- 1) = 3a-2b + C = 0, B = 0
f(-1)=-a+b-c=2
a=1；c=-3
f (x)=x^3-3x
2: Let g (x) = f (x) - f '(x) - 6x = x ^ 3-3x ^ 2-9x-3
Constant greater than m on [- 2,4]
Is 2 / 2 of the root a fraction,
If it is: √ 2 / 2 = √ 1 = 1, it is not a fraction
If it is: 2 / √ 2 = √ 2 / 2, it is irrational
In the interval [12,2], if the function f (x) = x2 + BX + C (B, C ∈ R) and G (x) = x2 + X + 1x have the same minimum value at the same point, then the maximum value of F (x) in the interval [12,2] is ()
A. 134B. 4C. 8D. 54
G (x) = x2 + X + 1 x = x + 1 x + 1 ≥ 3, if and only if x = 1, the equal sign holds, and the vertex coordinates of the function f (x) = x2 + BX + C are (1,3), x = − B2 = 11 + B + C = 3, then B = - 2, C = 4, f (x) = x2-2x + 4, f (x) max = f (2) = 4