F (x) = 2x * 3-3 (A-1) x * 2 + 1, find the monotone interval and extremum of the function

F (x) = 2x * 3-3 (A-1) x * 2 + 1, find the monotone interval and extremum of the function

f`(x)=6x^2-6(a-1)x
Let f '(x) = 0, x = 0, x = A-1
Discussion:
When a > 1, A-1 > 0, the increasing interval of F (x) is (- ∞, 0), (A-1, + ∞), and the decreasing interval is (0, A-1)
The maximum is f (0) = 1 and the minimum is f (A-1) = 1 - (A-1) ^ 3
When a
What is the domain? Is it symmetrical about the origin?
Be familiar with it
The domain is symmetric about the origin, that is to say, take any point x 1 in the domain, then - x 1 is also in the domain
For example, X ∈ [- 1,1], X ∈ (- ∞, - 2] or [2, + ∞) and so on
Let f (x) = x ^ 3 + x ^ 2-x, find monotone interval and extremum of function
F(X)=X^3+X^2-X
F ‘ (X) = 3X^2+2X-1 = (X+1)(3X-1)
When x ∈ (- ∞, - 1), it increases monotonically;
When x ∈ (- 1,1 / 3), it decreases monotonically;
When x ∈ (1 / 3, + ∞), it increases monotonically
Maximum f (- 1) = - 1 + 1 + 1 = 1
Minimum f (1 / 3) = 1 / 27 + 1 / 9-1 / 3 = - 5 / 27
Let f (x) = x ^ 3 + x ^ 2-x, find monotone interval and extremum of function
F’(x)=3x^2+2x-1
Let f '(x) = 0
3x^2+2x-1=0
(3x-1)(x+1)=0
x=1/3 x=-1
f’’(x)=6x+2
F '' (1 / 3) = 4 > 0 f (1 / 3) = - 5 / 27 minimum
f’’(-1)=-4
If the domain of the given function is not symmetric about the origin, then the function must be non odd and non even
Take Liezi for example
Take even function as an example: first, understand the definition of even function accurately: for any independent variable x in the domain, if f (- x) = f (x) holds, then the function f (x) is called even function. Second, take the function f (x) = x ^ 2, X ∈ (- 1,1] as an example. When x = 1, - x = - 1, although f (1) is solvable, but - 1 is not in the domain, f (- 1) is meaningless
Is this a difficult sentence? This means that if the domain of definition of a given function is not symmetrical about the origin, then the function must not be odd or even! There are many examples. F (x) = x, (0
Function f (x) = 1 / x + 2ln (x + 1) (1) find the monotone interval and extremum of F (x) (2) if f (x) is in
Function f (x) = 1 / x + AlN (x + 1)
(1) Finding monotone interval and extremum of F (x) when a = 4
(2) If f (x) is a monotone function in [2,4], find the value range of a, online, etc., urgent, no points, good people help
Wrong. When a = 2
First of all, the derivative function of question 1 is f '(x) = 2 / (x + 1) - (1 / x ^ 2) (Note: x ^ 2 is the square of x), let the derivative function f' (x) > 0, the scope of X is x1, because the definition domain of function is x > - 1. So the monotone increasing interval is (1, + ∞). The monotone decreasing interval is (- 1,1), and the minimum value is obtained at x = 1
How to determine the origin of trigonometric function
Why is x = 2K π + π / 2 not symmetric about the origin and x = 2K π is
When k = - K, x = - x, then x is symmetric about the origin
As above: x = 2K π + π / 2, let k = 1, then x = 5 / 2 π; when k = - 1, x = 3 / 2 π
So it's not symmetrical about the origin
Method 1: enumerate and calculate the number
Method 2: drawing and analytic geometry should be combined with figures and shapes
Method 3: if it is symmetric about the origin, then when k = - N, x = 2K π + π / 2 should be equal to - (- 2n π + π / 2). And it turns out that the equation doesn't hold, so it's not symmetrical about the origin.
And x = 2K π makes the equation hold, so it's symmetric. ... unfold
Method 1: enumerate and calculate the number
Method 2: drawing and analytic geometry should be combined with figures and shapes
Method 3: if it is symmetric about the origin, then when k = - N, x = 2K π + π / 2 should be equal to - (- 2n π + π / 2). And it turns out that the equation doesn't hold, so it's not symmetrical about the origin.
And x = 2K π makes the equation hold, so it's symmetric. Put it away
Let f (x) = x-2ln (x + 1), find the monotone interval and extremum of F (x)
The domain is x > - 1
f'(x)=1-2/(x+1)=(x+1-2)/(x+1)=(x-1)/(x+1)
The extreme point is x = 1
When - 1
How to judge the symmetry of the domain about the origin
Let's see if the distance between the two ends of the domain and the origin is equal. For example, the distance between [- 3,3], - 3 and 3 and the origin is equal, so they are symmetric
Urgent: given f (x) = x ^ 2-2ln ^ x, find the monotone interval and extremum of function f (x)
Let's make ln ^ x clear,
The definition field of F (x) = x ^ 2-2ln ^ x is {x | x > 0}
F '(x) = (x ^ 2-2ln ^ x)' = 2x-2 / x = 0, then x = 1
In (0,1), f '(x)
Judge the parity of the following functions: F (x) = 6-x, y = 2x + 2, f (x) = x, y = 3x + 1
F (- x) = x ^ 2 + 12x + 36 is neither equal to f (x) nor - f (x), so it is not odd or even
The second is that it means neither odd nor even
The third f (- x) = - x = - f (x) odd function
The fourth is also non odd and non even