Find the maximum and minimum values of the following functions in a given interval (1)f(x)=6x^2-x-2,x€[0,2] (2)f(x)=x^3-27x,x€[-4,4]

Find the maximum and minimum values of the following functions in a given interval (1)f(x)=6x^2-x-2,x€[0,2] (2)f(x)=x^3-27x,x€[-4,4]

(1) The derivative of F (x) = 6x ^ 2-x-2 is f '(x) = 12x-1, Let f' (x) = 012x-1 = 0x = 1 / 12, f (x) = - 49 / 24, x = 0, f (x) = - 2, x = 2, f (x) = 20, the maximum value is 20, the minimum value is - 49 / 24 (2) f '(x) = 3x ^ 2-27, when f' (x) = 0, 3x ^ 2-27 = 0, x = ± 3f (x) = x ^ 3-27x, x = ± 3, X =
Vector a = E1 + E2, B = 2e1-2e2, and E1 and E2 are collinear. Is a collinear with B? Let's give the reason
RT
Answer: A and B are collinear
Because E1 and E2 are collinear
So, there must be: E2 = k * E1 (k is a real number)
So,
a=e1+e2=（1+k)e1
b=2e1-2e2=2(1-k)e1= （2(1-k)/（1+k)）（（1+k)e1）
=（2(1-k)/（1+k)）a
So: A and B are collinear
Given the function f (x) = 4x-3x ^ 2, find the area s bounded by the image of (1) f (x) and the X axis of the tangent equation (2) f (x) at point x = 1
The slope k = f '(1) = - 2, y = - 2x + B. substituting (1, f (1) = 1) into the solution, the tangent equation y = - 2x + 3 of the image of B = 3, f (x) at point x = 1 is obtained. Let f (x) = 0, the area s = ∫ (0,4 / 3) (4x-3x & sup2;) DX = [2x & sup2; - X & sup3;] (0,4
1. For a project, Party A will complete it in X days, while Party B will complete it in y days, and the cooperation between the two parties is necessary_______ It will be finished in three days
If x = 6, y = 4, then two people cooperate______ It will be finished in three days
2. Some thermometers are marked with centigrade (° C) and Fahrenheit (° f), and the conversion formula between them is: C = 9 / 5 (f-32)
(1) Excuse me, can a person's temperature reach 100 ° f?
1. For a project, Party A will complete it in X days, while Party B will complete it in y days, and the cooperation between the two parties is necessary___ xy/(x+y)____ If x = 6, y = 4, then two people cooperate_ 2.4_____ 2. Some thermometers are marked with two temperature scales: centigrade (° C) and Fahrenheit (° f). The conversion formula between them is: C = 9 / 5 (f-32) (...)
xy/(x+y) 2.4
may not
1. The engineering quantity is 1.
（1） 1÷（1/x＋1/y）
(2) 1 ÷ (1 / 6 + 1 / 4) = 12 / 5 days
2. Can't reach
Decompose the following factors: X & # 178; - 4x + 8y-4y & # 178;
x²-4x+8y-4y²
=x²-4y²-4(x-2y)
=(x+2y)(x-2y)-4(x-2y)
=(x-2y)(x+2y-4)
solution
x²-4x+8y-4y²
=(x²-4y²)+(8y-4x)
=(x-2y)(x+2y)+4(2y-x)
=(x-2y)(x+2y)-4(x-2y)
=(x-2y)(x+2y-4)
x²-4x+8y-4y²
=（x+2y）(x-2y)-4(x-2y)
=(x-2y)(x+2y-4)
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x²-4x+8y-4y²
=（x+2y）(x-2y)-4(x-2y)
=(x-2y)(x+2y-4)
Hello, I'm glad to answer for you. Skyhunter 002 will answer your questions
If you don't understand this question, you can ask. If you are satisfied, remember to adopt it
If you have any other questions, please take this question and send it to me. Please understand. Thank you.
I wish you progress in your study
Find the monotone interval and extremum of cube-2 of function y = 3x-x, and the maximum and minimum values in the interval [0,4]
If f (x) = 3x-x & # 179; - 2, then f '(x) = 3-3x & # 178; = 3 (1-x) (1 + x)
Then the function f (x) decreases on (- ∞, - 1), increases on (- 1,1) and decreases on (1, + ∞). Then the minimum value of F (x) is f (- 1) = - 4 and the maximum value of F (x) is f (1) = 0;
In the interval [0,4], f (0) = - 2, f (1) = 0, f (4) = - 54. The maximum value of function f (x) in the interval [0,4] is f (1) = 0, and the minimum value is f (4) = - 54
First, we obtain the derivative, y '= 3-3x & # 178;, find the extreme point, x = ± 1, draw a picture, monotonically decrease in (- ∞, - 1) and (1, ∞), monotonically increase in [- 1,1], have the maximum value y = 0 when x = 1, and have the minimum value y = - 54 when x = 4.
Determine whether the vectors in the following questions are collinear: (1) a = 4e1-2 / 5e2, B = E1-1 / 10e2: (2) a = E1 + E2, B = 2E1_ 2e2, and E1, E2 are collinear
The condition is that E1 and E2 are not collinear, right?
(1) Collinear, a = 4B
(2) Not collinear, ax = B has no solution
On function: solving equations {2x-y = 3,3x + y = 7} by image method
The solution of the equation is the coordinates of the intersection of the functions y = 2x-3 and y = 7-3x
Because y is equal
So 2x-3 = 7-3x
So x = 2
So y = 2x-3 = 1
So the solution of the equation is x = 2, y = 1
1、(5x)\2-(1-2x)\0.2=5x 2、（4x-3）\0.5-5x-0.8=（1.2-x）\0.1
1、(5x)\2-(1-2x)\0.2=5x
5x-10(1-2x)=10x
5x-10+20x=10x
15x=10
x=2/3
2、（4x-3）\0.5-5x-0.8=（1.2-x）\0.1
8x-6-5x-0.8=12-10x
13x=18.8
x=94/65
Decomposition factor (a ^ 2 + 5A + 3) (a ^ 2 + 5a-2) - 6 x ^ 2 + 4xy + 4Y ^ 2 + 4x + 8y + 3 (x + 3) (x ^ 2-1) (x + 5) - 20
(a^2+5a+3)(a^2+5a-2)-6
x^2+4xy+4y^2+4x+8y+3
(x+3)(x^2-1)(x+5)-20
(a ^ 2 + 5A + 3) (a ^ 2 + 5a-2) - 6 let a & # 178; + 5A = a primitive = (a + 3) (A-2) - 6 = A & # 178; + a-6-6 = A & # 178; + A-12 = (a + 4) (A-3) and substitute a & # 178; + 5A = a into the primitive = (a + 4) (A-3) = (A & # 178; + 5A + 4) (A & # 178; + 5a-3) = (a + 1) (a + 4) (A & # 178; + 5a-3) = (a + 1) (a + 4) (A & # 178; + 5A