It is known that f (x) is an even function defined on R and is (0, positive infinity). Judge the monotonicity of F (x) on (negative infinity, 0) and prove it

It is known that f (x) is an even function defined on R and is (0, positive infinity). Judge the monotonicity of F (x) on (negative infinity, 0) and prove it

Even function images have opposite monotonicity with respect to the y-axis symmetric interval
Increasing on the left decreases on the right, decreasing on the left increases on the right
x> 0, f (- x) = f (x) because it is an even function
X0 f [- (- x)] = f (- x), that is, f (x) = f (- x)

If the domain of the function y = LG (AX ^ 2 + ax + 1) is the real number set R, find the value range of the real number a == So where did 4 come from = = II. I just can't calculate the binary quadratic equation. It's seven repairable.

If the domain of the function y = LG (AX ^ 2 + ax + 1) is a real number set R, then ax ²+ Ax + 1 > 0 is always true on R, which can be solved by using its image, but it should be discussed in terms of a = 0 and a ≠ 0. Answer: 0 ≤ a

Given that the definition field of function f (x) = LG (x ^ 2-ax + A / 2 + 2) is all real numbers, find the value range of A

x^2-ax+a/2+2=(x-a/2)^2+a/2+2-a^2/4
Because the domain of lgx must be x > 0, it holds for all real numbers when a / 2 + 2-A ^ 2 / 4 > 0
So we get a ^ 2-2a-8 (A-4) (a + 2) - 2

The function f (x) = x + X / a 1. If f (x) is an increasing function defined in the interval (0, + infinity), find the value range of real number a Solve the inequality f (2m-1) > F (m) under the condition of 1

f（x)=x+x/a = x(1+1/a)
Increasing function, 1 + 1 / a > 0
Get A0
f（2m-1)>f(m)
Because f is an increasing function
I.e. 2m-1 > m
Get m > 1

When x belongs to (2, positive infinity), the function y = LG (AX-1) is meaningful. Find the value range of real number a

For the function y = LG (AX-1) to be meaningful, it is necessary to make AX-1 > 0, then ax > 1. Because x > 2, if ax > 1 can be made when x takes the minimum value of 2, then ax > 1 can be made when x takes the greater value. Therefore, the answer is obtained by substituting 2 into x and solving a ≥ 1 / 2. (why can the equal sign be obtained here is because x cannot take 2, so a can be equal to 1 / 2.)

If the function f (x) = x + 1 / ax is an increasing function on (2, positive infinity), find the value range of real number a

f(x)=ax/（x+1）=（ax+a-a)/(x+1)=a-a/(x+1)
X + 1 is an increasing function
1 / (x + 1) is the subtraction function
-1 / (x + 1) is an increasing function
Therefore, it must be a > 0
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