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If the real numbers a and B satisfy the equation (A2 + B2 + 5) (A2 + b2-5) = 0, then A2 + B2=______
Using the square difference formula, let x = A2 + B2 > 0
The results are as follows:
(x+5)(x-5)=0
x2-25=0
x2=25
x = 5
therefore
a2+b2 =5
Given that the equation | x | = ax + 1 has a negative root but no positive root, the value range of a is ()
A. A ≥ 1b. A ＜ 1C. - 1 ＜ a ＜ 1D. A ＞ - 1 and a ≠ 0
∵ the equation | x | = ax + 1 has a negative root but no positive root, ∵ x ＜ 0, the equation is changed to: - x = ax + 1, X (a + 1) = - 1, x = − 1A + 1 ＜ 0, ∵ a + 1 ＞ 0, ∵ a ＞ - 1, and a ≠ 0. If x ＞ 0, | x | = x, x = ax + 1, x = 11 − a ＞ 0, then 1-A ＞ 0, the solution is a ＜ 1. ∵ a ＜ 1 is not established without positive root, ∵ a ＞ 1. Therefore, select a
Given that the equation | x | = ax + 1 has a negative root but no positive root, the value range of a is ()
A. A ≥ 1b. A ＜ 1C. - 1 ＜ a ＜ 1D. A ＞ - 1 and a ≠ 0
∵ the equation | x | = ax + 1 has a negative root but no positive root, ∵ x ＜ 0, the equation is changed to: - x = ax + 1, X (a + 1) = - 1, x = − 1A + 1 ＜ 0, ∵ a + 1 ＞ 0, ∵ a ＞ - 1, and a ≠ 0. If x ＞ 0, | x | = x, x = ax + 1, x = 11 − a ＞ 0, then 1-A ＞ 0, the solution is a ＜ 1. ∵ a ＜ 1 is not established without positive root, ∵ a ＞ 1. Therefore, select a
Monotone interval domain of function y = log base 2 (- x ^ 2 + 2x + 3)
Such as the title
Can't you do without drawing pictures?
Complex function problem
Domain: (- 1,3)
Process: (- x ^ 2 + 2x + 3) > 0
Range: (- infinity, 2]
Process: (- x ^ 2 + 2x + 3) greater than zero, and less than or equal to 4, y = log bottom 2, X image single increase, so bring in 0 and 4 to get the answer
Monotone interval: (- 1,1) minus (1,3) increasing
Process: composite function with the increase, difference and decrease, the two functions to draw a picture to know
The known function f (x) = - X & # 178; + 2x + 3
(1) Find the vertex coordinates of the function, the equation of the axis of symmetry; (2) find the singleton interval of the function; (if x belongs to [0,4], find the range of the function
The solution f (x) = - X & # 178; + 2x + 3 = - (X & # 178; - 2x) + 3 = - (X & # 178; - 2x + 1) + 4 = - (x-1) & # 178; + 4 vertex coordinates (1,4) axis of symmetry x = 1 the opening of the function is downward, the axis of symmetry is x = 1 when x1, the function y is a decreasing function the increasing interval is: (- ∞, 1) the decreasing interval is: (1, + ∞) x ∈ [0,4] when x = 1, the function
Find the domain, range and monotone interval of the function y = log (1 / 2) (- x ^ 2-2x + 8)
Domain: let - x ^ 2-2x + 8 > 0, that is, x ^ 2 + 2x-8
-X^2-2X+8=z>0 ( x+4)*(x-2)
Given function f (x) = (X & # 178; + 2x + 3) / X (x ∈ (2, + ∞))
(1) Find the minimum value of F (x) (2) if f (x) > A is constant, find the value range of A
Given the function y = (13) x2 + 2x + 5, find its monotone interval and range
Let t (x) = x2 + 2x + 5 = (x + 1) 2 + 4 ≥ 4, then the monotone decreasing interval of T (x) is (- ∞, - 1), the increasing interval is [- 1, + ∞) ∵ the function y = (13) t is the decreasing function, so the monotone increasing interval of function y = (13) x2 + 2x + 5 is (- ∞, - 1), the decreasing interval is [- 1, + ∞) ∵ 0 < y ≤ & nbsp; 181 ∵ the value range is (0181]
It is known that f (x) is a periodic function with the minimum positive period of 2 on R, and when 0 < x ≤ 2, f (x) = x3-2x2-x + 2, then the number of intersections between the image of function y = f (x) and X axis on interval [0,6] is ()
A. 6B. 7C. 8D. 9
When 0 ≤ x ＜ 2, Let f (x) = X3 - 2x2-x + 2 = 0 get x = 1 or x = 2, because f (x) is a periodic function with the minimum positive period of 2 on R, so f (1) = f (3) = f (5) = 0, f (0) = f (2) = f (4) = f (6) = 0, so the number of solutions of F (x) = 0 on interval [0,6] is 7, that is, the number of intersections of the image of function y = f (x) and X axis on interval [0,6] is 7, so B is selected
The sum of the exponents of the interval is known to be (2x + 5)
x²+2x+5=(x+1)²+4≥4
X-1 increment
And 0
x^2+2x+5=(x+1)^2+4>=4
Base 1 / 3 is greater than or less than 1
So (1 / 3) ^ x is a decreasing function
x^2+2x+5>=4
So (1 / 3) ^ (x ^ 2 + 2x + 5) 0
So the range (0,81)
Y = (1 / 3) ^ t is a decreasing function over the domain
T = x ^ 2 + 2x + 5 = (x + 1) ^ 2 + 4 where X-1 is an increasing function
So y = (1 / 3) ^ (x ^ 2 + 2x + 5) is minus X-1
Y
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