Given the function f (x) = - x ^ 3 + KX ^ 2 + 5x + 1, G (x) = - LNX + KX, where k ∈ R (1) when k = 1, find the extremum of the number of rows f (x)（ (2) If the equation f (x) = 0 with respect to X has a solution in the interval (1,2), find the value range of the real number K

Given the function f (x) = - x ^ 3 + KX ^ 2 + 5x + 1, G (x) = - LNX + KX, where k ∈ R (1) when k = 1, find the extremum of the number of rows f (x)（ (2) If the equation f (x) = 0 with respect to X has a solution in the interval (1,2), find the value range of the real number K

1）k=1,f(x)=-x^3+x^2+5x+1
F '(x) = - 3x ^ 2 + 2x + 5 = - (3x ^ 2-2x-5) = - (3x-5) (x + 1) = 0, the extreme point x = 5 / 3, - 1
F (- 1) = 1 + 1-5 + 1 = - 2 is the minimum
F (5 / 3) = - 125 / 27 + 25 / 9 + 25 / 3 + 1 = 202 / 27 is the maximum
2) From - x ^ 3 + KX ^ 2 + 5x + 1 = 0
Get: k = (x ^ 3-5x-1) / x ^ 2 = X-5 / X-1 / x ^ 2 = g (x)
In the interval (1,2), G '(x) = 1 + 5 / x ^ 2 + 2 / x ^ 3 > 0,
That is, G (x) increases monotonically, and the minimum value is g (1) = 1-5-1 = - 5
The maximum value is g (2) = 2-5 / 2-1 / 4 = - 3 / 4
So the value range of K is (- 5, - 3 / 4)
Given that the coordinates of points a, B and C are (- 2,1), (2, - 1), (0,1), CP = 3CA and CQ = 2CB respectively, the coordinates of point P, Q and vector PQ are obtained
CA = (- 2, 0), CB = (2, - 2), ｜ CP = 3CA = 3 (- 2, 0) = (- 6, 0), CQ = 2CB = (4, - 4). Let P (x, y). Then CP = (x, Y-1), ｜ x = - 6y-1 = 0, and the solution is x = - 6, y = 1.. P (- 6, 1). Similarly, q (4, - 3).. PQ = (10, - 4) can be obtained
On the distribution of real roots of quadratic equations with one variable
For several conditions of the distribution of real roots of quadratic equation of one variable, if the existence theorem of real roots is used, is it necessary to draw the inequality of discriminant
If one root of the real coefficient equation f (x) = x ^ 2 + ax + 2B = 0 is in (0,1), and the other root is in (1,2), find the range of (b-2) / (A-1) (using the method of simple linear programming)
For f (x) = 0, consider
Are the three inequalities f (0) > 0; f (1) < 0; f (2) > 0 sufficient
Is it necessary to add discriminant > 0
After adding the inequality about the discriminant, the feasible region changes from triangle to irregular figure, but it should be necessary
Does the real root existence theorem include the discriminant? If so, why is the feasible region obtained by the real root existence theorem different from the feasible region added with the discriminant?
Hope to answer
If f (0) > 0 and f (1) 0 are satisfied, the necessary discriminant is greater than 0, because such a quadratic function must have two intersections with the x-axis. As for the fact that the discriminant is not included in the graph, if the region of the graph changes after considering the discriminant, it is probably calculated or drawn
0.5x + 1 = 3, solution of linear equation with one variable,
0.5x=3-1
0.5x=2
X=4
0.5x+1=3
0.5x=2
X=4
Calculation: 1. (M + 1) (m ^ 4-m & # 179; + M & # 178; - M + 1); 2. (a + b) (a-2b) - (a + 2b) (a-b)
1.(m+1)(m^4-m³+m²-m+1)
=m^5-m^4+m³-m²+m+m^4-m³+m²-m+1
=m^5+1
2.(a+b)(a-2b)-(a+2b)(a-b)
=a²-2ab+ab-2b²-(a²-ab+2ab-2b²)
=a²-ab-2b²-a²-ab+2b²
=-2ab
1. The original formula = m (m ^ 4-m & # 179; + M & # 178; - M + 1) + (m ^ 4-m & # 179; + M & # 178; - M + 1) = m ^ 5 + 1
2. The original formula = (a ^ 2-ab-2b ^ 2) - (a ^ 2 + ab-2b ^ 2) = - 2Ab
One
m^5-m^4+m^3-m^2+m+m^4-m^3+m^2-m+1=m^5+1
Two
a^2-ab-2b^2-a^2-ab+2b^2=-2ab
Given the function f (x) = x ^ 3 - (k ^ 2-k + 1) x ^ 2 + 5x-2, G (x) = k ^ 2x ^ 2 + KX + 1, where k belongs to R, let P (x) = f (x) + G (x), if p (x) is in the interval（
The answer is "because P (x) is not monotone on (0 3), p '(x) = 0 has real solution and no multiple root on (0 3)"
Why not have heavy roots
If we have a quadratic function, then we have a quadratic function
The vertex is on the X axis
So the value of the function is always greater than or equal to or less than or equal to 0
So it's a monotone function
Given three points a (- 3, - 2) B (3,6) C (1,2), how to calculate the angle between Ca and CB of vector module
The results were arccos (- 3 / √ 10)
Discriminant of root of quadratic equation with one variable and its relation with coefficient
ax^2+bx+c=0
The discriminant of root △ = B ^ 2-4ac
If △ > 0, the equation has two different real solutions
If △ = 0, the equation has two identical real solutions
If △
△=B*B-4*A*C X1+X2= X1*X2=
B2-4ac > 0, 2. =0, 2 are the same, i.e. 1, less than 0, no root.
A determines the direction and size of the opening, a > 0 opening upward. a. B determines the axis of symmetry (- B / 2a), C determines the coordinates (0. C) of the y-axis
Is 2x + 1 = x-3 a linear equation with one variable
The answer is yes or no
2x+1=x-3
That is, x + 4 = 0 is a linear equation of one variable
1/2a²b³+M=1/2ab²（N+2b）
∵1/2a²b³=1/2ab²*N,
∴N=1/2a²b³÷1/2ab²=ab,
M=1/2ab²*2b=ab³.
This is how to find the expression of M and n
What do you want to ask?
1/2a²b³+M=1/2ab²（N+2b）
1/2a²b³+M-1/2ab²（N+2b）=0
1/2ab²(ab-N-2b)+M=0