If the quadratic function f (x) = x ^ 2 - (A-1) x + 5 is monotone in the interval (0,1), then the value range of a is

If the quadratic function f (x) = x ^ 2 - (A-1) x + 5 is monotone in the interval (0,1), then the value range of a is

Axis of symmetry x = (A-1) / 2
The quadratic function f (x) = x ^ 2 - (A-1) x + 5 is monotone in the interval (0,1),
So the axis of symmetry is not in the interval (0,1)
That is (A-1) / 2 = 1
a=3
Then the value range of a is
a=3

If the quadratic function f (x) = X2 - (A-1) x + 5 is an increasing function in the interval (12,1), the value range of F (2) is obtained

The quadratic function f (x) is an increasing function in the interval (12,1). Because the opening of its image (parabola) is upward, its axis of symmetry x = a − 12 or coincides with the straight line x = 12 or lies on the left side of the straight line x = 12, so a − 12 ≤ 12, the solution is a ≤ 2, so f (2) ≥ - 2 × 2 + 11 = 7, that is, f (2) ≥ 7

If the quadratic function f (x) = x ^ 2 - (A-1) x + 5 is an increasing function in the interval (0.5,1), find the value range of F (2)

The axis of symmetry of this function is x = (A-1) / 2
Because the function is an increasing function on (0.5,1), so
The axis of symmetry is on the left side of 0.5, i.e. (A-1) / 2

If the quadratic function f (x) = X2 - (A-1) x + 5 is an increasing function in the interval (12,1), the value range of F (2) is obtained

The quadratic function f (x) is an increasing function in the interval (12,1). Because the opening of its image (parabola) is upward, its axis of symmetry x = a − 12 or coincides with the straight line x = 12 or lies on the left side of the straight line x = 12, so a − 12 ≤ 12, the solution is a ≤ 2, so f (2) ≥ - 2 × 2 + 11 = 7, that is, f (2) ≥ 7

If the quadratic function f (x) - (A-1) x + 5 is an increasing function in the interval (0.5,1), then the value range of F (2) is

Because f (x) = X2 - (A-1) x + 5 is an increasing function in the interval {1 / 2,1}
So: F (1) - f (1 / 2) > 0
That is: [1 - (A-1) + 5] - {[(1 / 2) ^ 2] - (A-1) (1 / 2) + 5} > 0
The solution is: A6
That is: F (2) > 6

Let f (x) = x2-x + a (a > 0), if f (m) < 0, then the value of F (m-1) is ()
A. Positive number B. negative number C. non negative number D. positive number, negative number and zero are possible

Because the axis of symmetry of the function f (x) = x2-x + a (a > 0) is x = 12, and because a > 0, the approximate image of F (0) = a > 0 is as follows: from F (m) < 0 {0} m < 1 {M-1 < 0} f (m-1) > 0

Let f (x) = (M2 + 1) x2 + (m-1) x + 4 be the fixed-point coordinate of even function, which is the detailed process!

F (x) = (m ^ 2 + 1) x ^ 2 + (m-1) x + 4 is even function,
m=1,
f(x)=2x^+4,
Its vertex coordinates are (0,4)

If f (x) = (M2-1) x + (m-1) x + (n-2) is an odd function, then m n
Such as the title
process

F (x) = (M2-1) x ^ 2 + (m-1) x + (n-2) is an odd function
f(-x)=(m^2-1)x^2+(1-m)x+(n-2)=-f(x)
therefore
(m ^ 2-1) = (1-m ^ 2) = > m = 1 (rounding off) or, M = - 1
n-2=-(n-2) => n=2
Mainly use the function of the corresponding coefficient is equal to evaluate
m=-1,n=2

If f (x) = (m-1) x2 + (m-2) x + (m2-7m + 12) is even function, then the value of M is ()
A. 1B. 2C. 3D. 4

∵ function f (x) = (m-1) x2 + (m-2) x + (m2-7m + 12) is even function, ∵ f (- x) = f (x), ∵ m-1) X2 - (m-2) x + (m2-7m + 12) = (m-1) x2 + (m-2) x + (m2-7m + 12), ∵ m-2 = 0, M = 2, so B

The function f (x) = (m) is known_ 1)x^2+(m_ 2)x+(m^2_ 7 m + 12) is even function, then what is the value of M

∵ f (x) = (m-1) x & # 178; + (m-2) x + (M & # 178; - 7m + 12) are even functions
∴f(﹣x)＝f(x)
∴(m－1)x²－(m－2)x＋(m²－7m＋12)＝(m－1)x²＋(m－2)x＋(m²－7m＋12)
∴－(m－2)＝(m－2) ∴m＝2