Find the maximum and minimum values of the function f (x) = 1 / 3x3-x2-3x6 in the interval [- 2,5]

Find the maximum and minimum values of the function f (x) = 1 / 3x3-x2-3x6 in the interval [- 2,5]

f'(x)=x^2-2x-3
f'(x)=0 x^2-2x-3=0 (x-3)(x+1)=0
X = 3 or x = - 1
f(-2)=-8/3-4+6+6=16/3
f(-1)=-1/3-1+3+6=23/3
f(3)=9-9-9+6=-3
f(5)=125/3-25-15+5=20/3
Maximum = 23 / 3
Minimum = - 3

Given the function f (x) = (12 / x) + 3x (x > 0), then the minimum value of F (x) is,

F (x) = (12 / x) + 3x (x > 0) > = 2skr [(12 / x) * 3x] = 2skr36 = 12 (SQR = radical)

Minimum value of function f (x) = 3x + 12 / X (x) > 0

a^2+b^2>=2ab
Use this formula
(12/x)+3x>=12
When (12 / x) = 3x
When x = 2, the minimum value is y = 12

The minimum value of function f (x) = 3x + 12x2 (x > 0) is______ ．

∵ x > 0 ∵ 3x + 12x2 = 3x2 + 3x2 + 12x2 ≥ 333x2 · 3x2 · 12x2 = 9 if and only if 3x2 = 12x2, i.e. x = 2, the equal sign holds. Therefore, the minimum value of function f (x) = 3x + 12x2 (x > 0) is 9, so select: 9

If a nonzero vector a, B is known and a is perpendicular to B, then f (x) = (XA + b) & # is what function
A is an odd function and even function B is not odd or even C is odd function D is even function

The expansion is f (x) = | a | ^ 2 * x ^ 2 + 2A * b * x + | B | ^ 2,
Because a, B, a * b = 0,
So the function is f (x) = | a | ^ 2 * x ^ 2 + | B | ^ 2,
There is no linear term, so it is an even function
Choose D

It is known that the function f (x) = (1 / 3) x & # 178; - BX & # 178; + 2x + A, x = 2 is an extreme point of F (x)
It is known that the function f (x) = (1 / 3) x & # 178; - BX & # 178; + 2x + A, x = 2 is an extreme point of F (x)
Find the monotone interval of function f (x)

First, the derivative of F (x): F '(x) = (2 / 3-2b) x + 2;
If x = 2 is the extreme point, then f '(2) = 0, the solution is b = 5 / 6;
Then bring b back to f '(x) to calculate f' (x) > 0; f '(x)

Given the function f (x) = - x ^ 3 + ax ^ 2 + B, if the function f (x) has an extreme value at, and the minimum value is - 1, find the values of a and B
I'm mainly trapped in not knowing how to find out whether there is a minimum when x = 0 or x = 4

f(x)=-x³+ax²+b
f'(x)=-3x²+2ax
Let f '(x) = 0
-3x²+2ax=0
x(3x-2a)=0
X = 0 or x = 2A / 3
So f (x) has two stationary points,
If x = 0 and x = 4, we can get the extremum
So 2A / 3 = 4, so a = 6
f'(x)=-3x(x-4)
When f '(x) > 0, f (x) is a monotone increasing interval (0,4)
When f '(x)

Let X and y satisfy the condition 3x-y-6 = 0, x > = 0y > = 0. If the maximum value of the objective function z = ax + by (a > 0, b > 0) is 12, then the minimum value of 2 / A + 3 / B is?
The answer is 25 / 6
I know that the intersection of two straight lines is (4,6) ∧ 4A + 6B = 12, that is, a = (6-3b) / 2
Why is the solution of substituting a into 2 / A + 3 / b = 4 / (6-3b) + 3 / b > = 2 radical (12 / 6b-3b ^ 2) different from the answer?
It's a two radical [12 / (6b-3b ^ 2)]

>=The root of 2 (12 / 6b-3b ^ 2) is not very clear, and the maximum value may not be obtained
There's no need to bother
In this way, 4A + 6B = 12, so a / 3 + B / 2 = 1
The original formula 2 / A + 3 / b = (2 / A + 3 / b) (A / 3 + B / 2) = 2 / 3 + 3 / 2 + A / B + B / a > = 13 / 6 + 2 (A / b * B / a) = 25 / 6

Let X and y satisfy 3x-y-6 ≤ 0, X-Y + 2 ≥ 0 and X + y ≥ 0. If the maximum value of objective function z = ax + y (a > 0) is 14, then a = a.1 B.2 c.23 d.53/9

Y satisfies the constraint condition: 3x-y-6 ≤ 0, X-Y + 2 ≥ 0, X ≥ 0, y ≥ 0. If the objective function z = ax + X and the maximum value of Y, then x = 4Y = 6 can be obtained. Therefore, the problem becomes known as 4A + 6B = 1 and 2/

What is the relationship between the image of derivative function and that of original function

The image of derivative function is mainly drawn by using the relationship between derivative and function, so it mainly uses the relationship between the positive and negative of derivative and the monotonicity of function, the relationship between the maximum value and extreme value of function and derivative!