Any rational number can be represented by two points on the number axis. But we can't say what the points on the number axis only represent?

Any rational number can be represented by two points on the number axis. But we can't say what the points on the number axis only represent?

Fill in rational number
All the points on the number axis represent rational numbers______ (judge right or wrong)
The points on the number axis do not necessarily represent rational numbers. For example, the points on the number axis that represent π are not rational numbers, which is wrong
Every rational number can be represented by a point on the number axis, and the number represented by each point on the number axis may not be?
Every rational number can be represented by a point on the number axis, and the number represented by each point on the number axis is not necessarily rational
yes. Rational numbers can be represented by the number axis, but the points on the number axis can also represent rational numbers. Irrational numbers we have only studied π and E for the time being
It can also be irrational number, imaginary number, and irrational number measure is one, rational number measure is zero, that is, irrational number on the number axis is more than rational number.
Given the function f (x) = x ^ 3-3x + 2 (1), find the zero point of F (x); (2) find the value range of X satisfying f (x) 0 respectively
1.f(x)=x^3-3x+2=(x-1)^2*(x+2)
So when f (x) = 0, x = 1 or x = - 2
2. F (x) - 2 and X ≠ 1
Under the root sign (12 + under the root sign (12 + under the root sign (...) (infinite loop))
Let the answer of this formula be a. because the infinite cycle, the limit is the same, so under the root sign (12 + a) = a, the solution is a = 4
100% correct
Let's know the function f (x) = x ^ 3 + ax ^ 2 + BC + C, and let's take the maximum value 7 when x = - 1 and the minimum value when x = 3. Let's try to find the minimum value of function f (x) by finding the values of a, B and C
Given the function f (x) = x ^ 3 + ax ^ 2 + BX + C, we know that when x = - 1, we take the maximum value of 7, and when x = 3, we get the minimum value, so f '(x) = 3x ^ 2 + 2aX + B = 0, two are - 1,3 = = > - 1 + 3 = - 2A / 3, - 1 * 3 = B / 3, a = - 3, B = - 9 and - 1 + A-B + C = 7, C = 2, a = - 3, B = - 9, C = 2, f (3) = 27-27-27 + 2 = - 25
I tell you clearly, but the conditions, B and C, are not sure
Brother, you copied the wrong question. The original question is absolutely wrong. My estimate is f (x) = x ^ 3 + ax ^ 2 + BX + C
In this way:
f'(x)=3x^2+2ax+b
f'(-1)=3-2a+b=0
f'(3)=27+6a+b=0
f(-1)=-1+a-b+c=7
So a = - 3, B = - 9, C = 2 the minimum value is calculated by itself
Root 12 + root 8 times root 6
simple form
=√12+2√2×√6
=√12+2√12
=3√12
=6√3
=2 radical 3 + 4 radical 3 = 6 radical 3
Given that the function y = x ^ 3 + ax ^ 2 + BX + 27 has a maximum at x = 1 and a minimum at x = 3, what is a + B equal to
a＋b=3.
∵y'=3x^2＋2ax＋b
| y '| x = 1 = 3 + 2A + B = 0, similarly,
When y '| x = 3 = 27 + 6A + B = 0
The solution is a = - 6
I suggest you look at the chapter of derivative again
[root sign (three root signs, two)] can be expressed as?
=(3√2)^(1/2)
=3^(1/2)×2^(1/4)
If the image of the function f (x) = x * x * x-ax * x-bx is tangent to the X axis at the point (1,0), then f (x) has a maximum? A minimum?
Seeking maximum and minimum
f(x)=x^3-ax^2-bx
And X-axis tangent to point (1,0), then the function passes through point (1,0), and its derivative also passes through this point,
A = 2, B = - 1
So f (x) = x ^ 3-2x ^ 2 + X
If the derivative function is equal to 0, x = 1, x = 1 / 3
The maximum is 4 / 27 and the minimum is 0
Maxima and minima