Given that the definition field of function y = √ MX ^ 2-6mx + m + 8 is r, what is the value range of M? Analysis: 1 when m = 0, y = √ 8, the definition field is r 《2》 When m ≠ 0, make MX ^ 2-6mx + m + 8 ≥ + constant Only 1. M > 0 2.△=36m^2-4m(m+8)≤0 I.e. 0 ≤ m ≤ 1 Question 1 what is that triangle Question 2 how did 36m ^ 2-4m (M + 8) come from the formula? Because the foundation is poor, I don't understand much. Please explain the answer to this question in detail

Given that the definition field of function y = √ MX ^ 2-6mx + m + 8 is r, what is the value range of M? Analysis: 1 when m = 0, y = √ 8, the definition field is r 《2》 When m ≠ 0, make MX ^ 2-6mx + m + 8 ≥ + constant Only 1. M > 0 2.△=36m^2-4m(m+8)≤0 I.e. 0 ≤ m ≤ 1 Question 1 what is that triangle Question 2 how did 36m ^ 2-4m (M + 8) come from the formula? Because the foundation is poor, I don't understand much. Please explain the answer to this question in detail

The triangle is the square of the discriminant ax + BX + C and the square of B - 4ac
36m ^ 2-4m (M + 8) is calculated by the above discriminant

The process of finding the higher derivative of y = arctanx and y = arcsinx should be detailed To the third order

y=arctanx
y'=1/(1+x ²)
y''=-2x/(1+x ²)²
y'''=(6x ²- 2)/(x ²+ 1) ³
y=arcsinx
y'=1/(1-x ²)^ (1/2)
y''=x/(1-x ²)^ (3/2)
y'''=(2x ²+ 1)/(1-x ²)^ (5/2)

F (x) = (- x square + ax) the x-square of E. what is the derivative of this function? It is proved that the absolute value of the two extreme values x1-x2 is greater than or equal to 2

f(x)=(-x2+ax)e^x
f'=(-x2+ax)'·e^x+(-x2+ax)·(e^x)'
=(-2x+a)e^x+(-x2+ax)e^x
=(-x^2-2x+ax+a)e^x

The monotonicity of the function f (x) = x + 2 / X on the interval {- radical 2, 0) is proved Such as the title

Direct definition
Set - root 2 "X1"

The monotonicity of the function f (x) = 1 / radical 1-2x is judged and proved

The function is an increasing function. It is proved as follows: firstly, calculate the definition field of the function, and it can be obtained from √ (1-2x) as the denominator: 1-2x > 0, that is, x < 1 / 2. In (- ∞, 1 / 2), let X1 < x2 < 1 / 2, f (x2) - f (x1) = 1 / √ (1-2x2) - 1 / √ (1-2x1) = [(√ (1-2x1) -√ (1-2x2)] / √ [(1-2x1) (1

——Judge and prove the monotonicity of function f (x) = √ 1-x on interval (negative infinity, 1). How to remove the root sign

Set x1