F (x) = 1 / 3x3 ln (2 + x3), find the maximum and minimum of the function

F (x) = 1 / 3x3 ln (2 + x3), find the maximum and minimum of the function

F '(x) = x ^ 2 - (3x ^ 2) / (2 + x ^ 3) = x ^ 2 (x ^ 3-1) / (2 + x ^ 3) = 0, the extreme point x = 0,1
f'(0+)0,f'(1-)

For the function f (x) = x3-x2 + X, the following statement is correct ()
A. There are maxima and no minima B. There are minima and no minima C. There are both maxima and minima d. There are neither maxima nor minima

∵ function f (x) = x3-x2 + X, ∵ f ′ (x) = 3x2-2x + 1, Let f ′ (x) = 0, ∵ f ′ (x) = (3x + 1) (x-1) = 0, the solution is x = 1 or - 13, when f ′ (x) > 0, we can get x ＞ 1 or X ＜ - 13, then the increasing function of F (x); when f ′ (x) ＜ 0, we can get - 13 ＜ x ＜ 1, then the decreasing function of F (x); the minimum point of F (x) is at x = 1, and the maximum point of F (x) is at x = - 13, Therefore, C is chosen;

Let f (x) = Xe ^ x, a.x = 1 be the maximum point of F (x), b.x = 1 be the minimum point of F (x), c.x = - 1 be the maximum point of F (x), D.X = - 1 be the minimum point of F (x)

F '(x) = e ^ x + Xe ^ x = (1 + x) e ^ x = 0, the extreme point x = - 1
f"(x)=e^x+(1+x)e^x=(2+x)e^x
f"(-1)=e^(-1)>0
So x = - 1 is the minimum point
Choose D

The maximum value of function f (x) = x ^ 4-x ^ 2 is

f'(x)=4x³-2x
Let f '(x) > 0, then: - √ 2 / 2

The maximum point of function f (x) = (x + 2) 2 (x-1) 3 is ()
A. X = - 2 or 1b. X = - 1 or 2C. X = - 1D. X = - 2

∵ f ′ (x) = (X-2) (X-2) 2 (5x + 4), Let f ′ (x) > 0, the solution is: X ＞ - 45, or X ＜ - 2, ∵ function f (x) increases on (- ∞ - 2), (− 45, + ∞), decreases on (- 2, - 45), ∵ x = - 2 is the maximum point of the function, so we choose; D

The minimum value of function y = 2x ^ 3-6x ^ 2-18x + 7 is?

y'=6x²-12x-18=0
x=3,x=-1
X3, y '> 0, increasing
-1

Y = 2x ^ 3-6x ^ 2-18x + 7 to find the extremum of function

y'=6x^2-12x-18=0
x^2-2x-3=0
（x-3）（x+1）=0
x=-1,x=3
Maximum y = (- 1) = 17,
Minimum y = (3) = - 47

7.1 determine the monotone interval of function y = 2x ^ 3-6x ^ 2-18x-7

y' = 6X^2 - 12X - 18
Let y '= 0
(X + 1)(X - 3) = 0
X = - 1 or x = 3
When x > = 3 or X

What is the increasing interval of function y = 2x ^ 3-6x ^ 2-18x-7?

Derivation: y '= 6x ^ 2-12x-18
Let y '= 0, that is, 6x ^ 2-12x-18 = 0, the solution is: X1 = - 1, X2 = 3
When X1 = 3, the value of derivative function y 'is greater than or equal to 0, and the function increases monotonically

Given the function f (x) = f (x) = - x ^ 3 + 6x ^ 2 + 9x + 7
1. Find the definition field of function
2. Monotone interval and extremum of function
3. Function concave convex interval and inflection point

I really hate these questions. It's useless after the exam. Now I forget all of them