Given the function f (x) = x ^ 3 + BX + CX + 2, the extremum is obtained at x = 2 / 3 Determine the analytic expression of function f (x) and find the monotone interval of function f (x)

Given the function f (x) = x ^ 3 + BX + CX + 2, the extremum is obtained at x = 2 / 3 Determine the analytic expression of function f (x) and find the monotone interval of function f (x)

First, we get f '(x) = 3x ^ 2 + B + C
When f '(2 / 3) = 0, the solution is B + C = - 4 / 3
So f (x) = x ^ 3 - (4 / 3) * x + 2
If the monotone increasing interval f '(x) is greater than 0, the inequality can be solved. X is greater than 2 / 3 or X is less than - 2 / 3
If the derivative is less than 0, the inequality is solved. X is less than 2 / 3 and X is greater than - 2 / 3
(the equal sign can be taken or not, without too many requirements!)

Let f (x) = ax & sup3; + BX & sup2; + CX have extremum at x = 1 and x = - 1, and f (1) = - 1. Find the analytic expression of F (x) of the function

f（x）＝ax³＋bx²＋cx
f'（x）＝3ax^2＋2bx＋c=0
Substitute x = 1 and x = - 1 to get
3a+2b+c=0 (1)
3a-2b+c=0 (2)
f（1）＝－1
Namely
-1=a+b+c (3)
From the system of equations
b=0,a=1/2,c=-3/2
f(x)=1/2x^3-3/2x

A cube box with an edge length of 4 decimeters has a surface area of () square decimeters. If the top cover is removed, what is the surface area of the uncovered cube box?

The surface area of a cube box with a length of 4 decimeters is (96) square decimeters. If the top cover is removed, the surface area of the uncovered cube box is (80) square decimeters

If the equation x2-mx + 2 = 0 and X2 - (M + 1) x + M = 0 have the same real root, then the value of M is ()
A. 3B. 2C. 4D. -3

From the equation x2-mx + 2 = 0 to get x2 = mx-2, from the equation X2 - (M + 1) x + M = 0 to get x2 = (M + 1) x-m. then there is mx-2 = (M + 1) x-m, that is, x = m-2. Substituting x = m-2 into the equation x2-mx + 2 = 0 to get the equation (m-2) 2-m (m-2) + 2 = 0, then the solution is m = 3

If | A-2 | and | B + 5 | are opposite to each other, find the absolute value of a + B

|a-2|+|b+5|=0
|a-2|=0
|b+5|=0
a-2=0
a=2
b+5=0
b=-5
|a+b|=|2-5|=3

The domain of the function f (x) = LG (sin2x cos2x) is______ ．

From the meaning of the question, we can get: sin2x-cos2x ＞ 0, that is, cos2x-sin2x ＜ 0. From the double angle formula, we can get cos2x ＜ 0, so π 2 + 2K π ＜ 2x ＜ 3 π 2 + 2K π, K ∈ Z, ∧ K π + π 4 ＜ x ＜ K π + 3 π 4, K ∈ Z, so the answer is: {x | K π + π 4 ＜ x ＜ K π + 3 π 4, K ∈ Z}

If the absolute value of a + the absolute value of B = the absolute value of a + B, find the conditions that a and B satisfy

Because | a | + | B | = | a + B |,
Square a ^ 2 + B ^ 2 + 2 | ab | = a ^ 2 + B ^ 2 + 2Ab,
So | ab | = AB,
This shows that ab > = 0, that is, a and B have the same sign or 0 in them

The known points a (a, Y1), B (2a, Y2), C (3a, Y3) are all on the parabola y = 5x & sup2; + 12x
1) Finding the coordinates of the intersection of parabola and X-axis
(2) When a = 1, find the area of triangle ABC

1)y=5x^2+12x=0
Coordinates of intersection point with X axis (0,0), (- 12 / 5,0)
2)a=1
y1=17,y2=44,y3=81
A(1,17)B(2,44),C(3,81)
AB=√730
Straight line AB: 27x-y-10 = 0,
C to ab distance D (high)
D=|27*3-1*81-10|/√730
Area of triangle ABC
=1/2*D*AB=5

Function y =! (x-1) Note: the absolute value can't be typed with! Instead of whether there is a derivative at X. = 0? If so, find out its derivative
Function y =! (x-1) Note: the absolute value can not be typed! Instead of whether there is a derivative at X. = 0? If so, find out its derivative. Analytically, if the function is differentiable, then the left and right limits at the inflection point on the image are the same, that is, the left derivative equals the right derivative. How to do this kind of problem? I want to understand that this problem is not the solution steps of this problem

If the function is differentiable if and only if the left derivative is equal to the right derivative
When x tends to zero, Lim [f (0 + △ x) - f (0)] / △ x = 1
When x tends to zero, Lim [f (0 + △ x) - f (0)] / △ x = - 1
So there is no derivative at x = 0

The properties and images of the function f (x) = 1 + 2x / x2
X belongs to R

F（x）=2x/（1+x²）
F (- x) = - 2x / (1 + X & # 178;) we know that it is an odd function
F（0）=0
When x > 0
F (x) = 2x / (1 + X & # 178;) numerator denominator divided by X
          =2/（1/x+x）
The molecule is a pair of hook functions decreasing at (0,1) and increasing at (1, + ∞)
The f (x) increases in (0,1) and decreases in (1, + ∞)
Combined with the x < 0 part
It can be seen that f (x) decreases in (- ∞, - 1) and [1, + ∞]
In [- 1,1]
But when x ＜ 0, f (x) ＜ 0. When x ＞ 0, f (x) ＞ 0