The derivative function f '(x) = a (x + 1) (x-a) of function f (x) is known. If f (x) takes a minimum at x = a, the value range of real number a is __

From F ′ (x) = a (x + 1) (x-a) = 0, the solution is a = 0 or x = - 1 or x = A. if a = 0, then f ′ (x) = 0. At this time, the function f (x) is a constant and has no extreme value, so a ≠ 0. If a = - 1, then f ′ (x) = - (x + 1) 2 ≤ 0. At this time, the function f (x) decreases monotonically and has no extreme value, so a ≠ - 1. If a ＜ - 1, from F

Given that the value range of the function f (x) = (MX ^ 2 + 8x + n) / x ^ 2 + 1 is [1,9], find the values of M and N? Answer (last steps): From 1 ≤ y ≤ 9, the two of the univariate quadratic equation y ^ 2 - (M + n) y + mn-16 = 0 are 1 and 9 So m + n = 1 + 9, mn-16 = 1 × nine m=n=5

Answer: F (x) = y = (MX) ²+ 8x+n)/(x ²+ 1)yx ²+ y=mx ²+ 8x + n finishing: (y-m) x ²- 8x + y-n = 0. This is a univariate quadratic equation about X in the real number range R. there is always a real number solution. Therefore, the discriminant = (- 8) ²- 4 * (y-m) * (y-n) > = 0: y ²- (m+n)y+...

Given that the definition field of the function f (x) = (MX square + 8x + n) / x square + 1 is R and the value field is [1,9], find the value of M and n

Let t = (MX2 + 8x + n) / (x2 + 1) then 1=

Given that the value range of the function f (x) = (MX ^ 2 + 8x + n) / (x ^ 2 + 1) is [1,9], X ∈ R, find the values of M and N?

It's not easy for me to come up with a method, but it can be calculated. You can practice the formula yourself. It's a little too much
1. Find the derivative of F (x) = (MX ^ 2 + 8x + n) / (x ^ 2 + 1), and obtain the two relations between two X1 x2 and m n with Weida theorem for molecules, which are recorded as Formula 1 and formula 2. At the same time, write the discriminant of molecules to test the obtained m n
2. Substituting X1 and X2 into f (x) respectively, the other two formulas are equal to the two maximum values, which are recorded as formula 3 and formula 4
3.1 equations 2, 3 and 4 are combined to establish four unknowns and four equations. The equations can be solved and may have additional roots. Remember to use the discriminant test. Finish the answer

It is known that the definition field of function y = log3 ((MX ^ 2 + 8x + n) / (x ^ 2 + 1) is R and the value field is (0)

The solution of MX ^ 2 + 8x + n > 0 is x ∈ R (obviously m ≠ 0)
m>0
eight ² - 4mn>=0 (1)
M < 0, that's impossible
0=0 (2)
(m-9)x ² + 8x + (n -9)m

Known function f (x) = x2-4x + A + 3, G (x) = MX + 5-2m (I) if y = f (x) has zero on [- 1,1], find the value range of real number a; (II) when a = 0, if there is always x2 ∈ [1,4] for any x1 ∈ [1,4], make f (x1) = g (x2), and find the value range of the real number m; (III) if the range of the function y = f (x) (x ∈ [T, 4]) is interval D, is there a constant t so that the length of interval D is 7-2t? If it exists, find all the values of T; If not, please explain the reason (Note: the length of interval [P, q] is Q-P)

The function f (x) = x2-4x + A + 3, G (x) = MX + 5-2m. (I) if y = f (x) has zero on [- 1,1], find the value range of real number a; (II) when a = 0, if there is always x2 ∈ [1,4] for any x1 ∈ [1,4], make f (x1) = g (x2), and find the value range of the real number m; (III) if function y =

The derivative function f '(x) = a (x + 1) (x-a) of function f (x) is known. If f (x) takes a minimum at x = a, the value range of real number a is __