The monotonicity interval (1) f (x) = x Λ 2 + 2X-4 (2) f (x) = 2x Λ 2-3x + 3 (3) f (x) = 3x + X Λ 3 is obtained （4）f(x)=x∧3+x∧2-x

The monotonicity interval (1) f (x) = x Λ 2 + 2X-4 (2) f (x) = 2x Λ 2-3x + 3 (3) f (x) = 3x + X Λ 3 is obtained （4）f(x)=x∧3+x∧2-x

(1) If f '(x) = 2x + 2, Let f' (x) = 0 get x = - 1, when x ＞ - 1, f '(x) ＞ 0, when x ＜ - 1, f' (x) ＜ 0, so decrease interval (- ∞, - 1), increase interval (- 1, + ∞) (2) f '(x) = 4x-3, Let f' (x) = 0, get x = 3 / 4, when x ＞ 3 / 4, f '(x) ＞ 0, when x ＜ 3 / 4, f' (x) ＜ 0, so decrease
How is the domain of F (x) symmetric about the origin?
The solution draws the domain of the function on the number axis,
If the area drawn is symmetrical about the origin on the number axis,
be
The domain of the function f (x) is symmetric with respect to the origin
Given the function f (x) = 1 / 3x ^ 3-3 / 2x ^ 2 + 2x + 1, find the monotone interval of F (x)
The general way to find the interval of higher order function is to find the derivative
f'(x)=x^2-3x+2
Then the interval of F '> - 0 is the increasing interval of the original function
How to know if the domain of definition is origin symmetry and y-axis symmetry? For example, why the domain of definition of even function f (x) is (T, 2t-3) origin symmetry
That is to say, every x in the function has a corresponding - x corresponding to it, otherwise it does not conform to the definition of odd even function. When the domain of definition of even function f (x) is (T, 2t-3), it must be that T and 2t-3 are two opposite numbers to satisfy the symmetry on both sides of the origin, and T "0,2t-3" 0
If not, the domain itself is not symmetric about the origin
Given that the tangent of function f (x) = x3-3x2 + ax + B at x = - 1 is parallel to X axis, (1) find the value of a and the monotone interval of function f (x); (2) if the image of function y = f (x) and parabola y = 32x2-15x + 3 have exactly three different intersections, find the value range of B
(1) From the known f ′ (x) = 3x2-6x + A, the tangent at x = - 1 is parallel to the X axis, and the solution is a = - 9. Then f ′ (x) = 3x2-6x-9 = 3 (x + 1) (x-3). From F ′ (x) > 0, the solution is x ＞ 3 or X ＜ - 1. From F ′ (x) ＜ 0, the solution - 1 ＜ x ＜ 3
If the domain is symmetric about the origin, does the function have parity
not always
such as
F (x) = x (when x0)
This function is defined on R. but it has no parity
F (x) = 1 / 3x ^ 3 + x ^ 2-ax. If a = 3, find the monotone interval of F (x)
F (x) = (1 / 3) x ^ 3 + x ^ 2-3x, Let f '(x) = x ^ 2 + 2x-3 = (x + 3) (x-1)
f(x)=1/3x^3+x^2-ax
f'(x) = x^2+2x-a = (x+1)^2+1-a
When a ≤ 1,
F '(x) ≥ 0, f (x) increases monotonically on R
That is, monotone increasing interval (- ∞, + ∞)
When a > 1,
f'(x) = x^2+2x-a = (x+1)^2-(a-1) = {x+1+√(a-1)}{x+1-√(a-1)}
Monotone increasing interval (- ∞, - 1 - √ (a... expansion)
f(x)=1/3x^3+x^2-ax
f'(x) = x^2+2x-a = (x+1)^2+1-a
When a ≤ 1,
F '(x) ≥ 0, f (x) increases monotonically on R
That is, monotone increasing interval (- ∞, + ∞)
When a > 1,
f'(x) = x^2+2x-a = (x+1)^2-(a-1) = {x+1+√(a-1)}{x+1-√(a-1)}
Monotone increasing intervals (- ∞, - 1 - √ (A-1)) and (- 1 + √ (A-1), + ∞)
Monotone decreasing interval (- 1 - √ (A-1), - 1 + √ (A-1)) is closed
What is "domain of definition must be symmetric about origin"
Please explain clearly. I'm a freshman. I don't understand
If the domain is (- 1,1), (- 2,2), (- 3,3), [- 4,4], then the domain is symmetric with respect to the origin. Note that the domain is also symmetric with respect to R (real number)
(- 1,1] asymmetry, because an open interval, a closed interval
In other words, for any set, the definition of - D ∈ - is symmetric
It means that this function is odd. We have the formula F (x) = - f (- x)
The monotone interval of function f (x) = x ^ 3 - 3x ^ 2 + 1 is
I have derived: F '(x) = 2x ^ 2 - 6x
x,= 0 x,= 3
But I still can't figure out the final answer-
Halo, derivative is wrong
f(x) = x^3 - 3x^2 + 1
f'(x)=3x^2-6x
Let f '(x) = 0
3x^2-6x=0
X = 0 or x = 2
When x
What does domain of definition mean about origin symmetry?
First of all, if the domain is drawn on the number axis, it may be some isolated points, not necessarily line segments or rays
The domain of definition is symmetric with respect to the origin. 1. There is nothing to be seen in the function image. 2. It is a necessary condition for a function to have an odd or even function. It is impossible to leave it, but only satisfy it, and it is not necessarily an odd or even function
3. You know the three elements of a function. Only the function formula, for example, y = (x-1) &# 178;, if we define its domain of definition as [- 3,3], then the domain of definition is symmetric about the origin, but the image is not symmetric about X, y, O points
4. For another example, the definition field of logarithmic function is basically [negative number and zero have no logarithm]. Therefore, in order to make the definition field of [logarithmic function type] symmetric about O, we have to design it as follows: y = ㏒ 3 (X & # 178; - 2)
Don't you know what I'm saying?