If function f (x)= a2-1x2+（a-1）x+2 The definition field of a + 1 is r, and find the value range of A

If function f (x)= a2-1x2+（a-1）x+2 The definition field of a + 1 is r, and find the value range of A

∵ function f (x)=
a2-1x2+（a-1）x+2
The domain of a + 1 is r,
‡ a satisfied
a2-1≥0
a+1≠0 ，
Namely
A ≥ 1 or a ≤ - 1
a≠-1 ，
Then a ≥ 1 or a ＜ - 1

Given that the definition field of x ^ 2 + X + A under the function f (x) = root sign is r, find the value range of real number a The answer is a > = 1 / 4

If the domain is r, then x ²+ X + a > = 0
x ²+ x+1/4-1/4+a
=(x+1/2) ²- 1/4+a
(x+1/2) ²>= 0
So as long as - 1 / 4 + a > = 0
a>=1/4

If the definition field of the root opening sign of the function {(a ^ 2-1) - (A-1) x + 2 / (a + 1)} is r, find the value range of the real number a

Root sign ((a ^ 2-1) - (A-1) x + 2 / (a + 1))
Because the domain is r
Therefore, A-1 = 0
a=1

Given that the definition field of function f (x) = 1 / root sign (MX ^ 2 + 4mx + 3) is r, then the value range of M

The definition field is r, that is, MX ^ 2 + 4mx + 3 = 0 has no solution
That is, B ^ 2-4ac

Given that the definition field of the function y = root sign (square of MX - 6mx + m + 8) is r, then the value range of M is. For example, why △

It is known that the domain of the function is r
Therefore, when x takes any real value, it is meaningful in the root sign, that is, the quadratic polynomial in the root sign must be ≥ 0
Therefore, the intersection of the image of the function and the x-axis can only have one intersection (△ = 0) or no intersection (△)

Known function y= The domain of MX2 − 6mx + m + 8 is r (1) Find the value range of real number m; (2) When m changes, if the minimum value of Y is f (m), find the value range of function f (m)

(1) According to the meaning of the topic, when x ∈ R, mx2-6mx + m + 8 ≥ 0 is always true. When m = 0, X ∈ R;
When m ≠ 0,
m>0
△≤0
Namely
m>0
(-6m)2-4m(m+8)≤0 ．
The solution is 0 < m ≤ 1, so the value range of real number m is 0 ≤ m ≤ 1
(2) When m = 0, y = 2
2；
When 0 < m ≤ 1, y=
m(x-3)2+8-8m．
∴ymin=
8-8m．
Therefore, f (m)=
8-8m（0≤m≤1），
Easy to get 0 ≤ 8-8m ≤ 8
The value range of F (m) is [0,2]
2]．