If y = xsquare - 2mx + 1 is a decreasing function on (negative infinity, 2], then the range of M is

If y = xsquare - 2mx + 1 is a decreasing function on (negative infinity, 2], then the range of M is

M is greater than or equal to 2
The axis of symmetry of this function is x = M. if the function is a decreasing function on (- ∞, 2), then x = m is on the line or right side of x = 2. Therefore, M is greater than or equal to 2
{x2+x x>0 f(x)={ {-x2+x x
Judging the parity of function
{x2+x x>0
f(x)= {
{-x2+x x
When x > 0, - x
If f (x) = the square of X + 2mx + 2 is a decreasing function in the range of less than or equal to 1, find the value range of M
F (x) = the square of X + 2mx + 2 is a decreasing function in the range of less than or equal to 1
The opening of the image is upward, so the axis of symmetry x > = 1
x=-m>=1
M
Judging the parity of function: F (x) = x2 + a △ x.x ≠ 0
F (- x) = - X & # 178; + A / (- x) = x & # 178; - A / X if a function is even, then f (- x) - f (x) = 0 if a function is odd, then f (- x) + F (x) = 0 ∵ f (- x) - f (x) = (X & # 178; - A / x) - (X & # 178; + A / x) = - 2A / x, f (- x) + F (x) = (X & # 178; - A / x) + (X & # 178; + A / x) = 2x & # 178
The function f (x) = x & # 178; + 2 (A-1) x + 2 is a decreasing function in the interval (- ∝, 4), then the value range of a is?
The opening is upward, and the left side of the axis of symmetry is decreasing,
The symmetry axis is x = 1-A, and the interval (- ∞, 4) is on the left side of the symmetry axis X = 1-A,
Then: 4 ≤ 1-A
a≦-3
The function equation is regarded as a parabola, the abscissa of the vertex is x = A-1 and the opening is upward
Therefore, on the left side of x = A-1, it decreases monotonically
So A-1 ≥ 4  a ≥ 5
The definition domain of function f (x) = root sign 1-x / 3x square + 1g (3x + 1) is a. (- 1 / 3, + ∞) B. (- 1 / 3)
The definition field of function f (x) = root sign 1-x / 3x square + 1g (3x + 1) is
A. (- 1 / 3, + ∞)
B. (- 1 / 3, 1)
C. (- 1,3]
D. (- ∞, - 1 / 3)
B. (- 1 / 3, 1)
If the function f (x) = x & # 178; + 2 (A-1) x + 2 is a decreasing function in the interval (- infinite, 4), then the value range of a is
The symmetry axis - (2 (A-1)) / 2 = 1-A > = 4, and a is obtained
LIM (x tends to 0) (tanx-x) / (x-sinx)
LIM (x tends to 0) (tanx-x) / X & # 179; = LIM (x tends to 0) (SEC & # 178; x-1) / (3x & # 178;) = LIM (x tends to 0) (1-cos & # 178; x) / (3x & # 178; Cos & # 178; x)
=LIM (x tends to 0) (SiNx / x) &# 178; / (3cos & # 178; x) = 1 / 3
LIM (x tends to 0) x & # 179; / (x-sinx) = LIM (x tends to 0) 3x & # 178; / (1-cosx) = LIM (x tends to 0) 3x & # 178; / 2 Sin & # 178; (x / 2)
=LIM (x tends to 0) 6 [(x / 2) / sin (x / 2)] & #178; = 6
Original limit = (1 / 3) 6 = 2
Using the law of Robida
Original formula = LIM (x - > 0) (SEC ^ 2x-1) / (1-cosx)
=lim(x->0) (2sec^2xtanx)/(sinx)
=2*lim(x->0) 1/cos^3x
=2
(1) If the quadratic function f (x) = x & # 178; + 2 (A-1) is a decreasing function in (- ∞, 1), then the value range of a is
(2) The range of the function y = 1 / sqrt (X & # 178; + 1),
The first question is wrong
2(a-1)x?
The opening is upward and the axis of symmetry decreases to the left
So x = - (A-1) ≥ 1
a-1≤0
a≤1
x²+1>=1
√(x²+1)>=1
So 0
（1）f（x)=x²+2（a-1)
If the image opening of the quadratic function is upward and the axis of symmetry is x = - (A-1) = 1-A, then the subtraction interval of the function is (- ∞, 1-A)
It is also known that the quadratic function f (x) = x & # 178; + 2 (A-1) is a decreasing function in (- ∞, 1)
So the subtraction interval of the original function is larger than the known interval,
So: 1-A should be on the right side of 1.
So 1-A > = 1, a = 1, a
LIM (x tends to 0) (TaNx SiNx) / x ^ 3 =?
lim(x→0)(tanx-sinx)/x^3
=lim(x→0)(sinx/cosx-sinx)/x^3
=lim(x→0)(1-cosx)/x^2
=lim(x→0)1/2x^2/x^2
=1/2
Zero point five