It is known that the function f (x) = 3x, f (a + 2) = 18, G (x) = λ· 3ax-4x has the semantic field of [0,1]. (I) find the value of a; (II) if the function g (x) is a monotone decreasing function in the interval [0,1], find the value range of real number λ

It is known that the function f (x) = 3x, f (a + 2) = 18, G (x) = λ· 3ax-4x has the semantic field of [0,1]. (I) find the value of a; (II) if the function g (x) is a monotone decreasing function in the interval [0,1], find the value range of real number λ

(I) it is known that 3A + 2 = 18 {3A = 2} a = log32 (II) in this case, G (x) = λ· 2x-4x, let 0 ≤ X1 < x2 ≤ 1, because g (x) is a monotone decreasing function in the interval [0, 1], so G (x1) - G (x2) = (2x2-2x1) (- λ + 2x2 + 2x1) ≥ 0 holds ∵ 2x2-2x1 > 0 ∵ λ ≤ 2x2 + 2x1 holds. Because 2x2 + 2x1 ≥ 20 + 20 = 2, the value range of real number λ is λ ≤ 2

Let f (x) = log3 (x ^ 2 + ax + b) / (x ^ 2 + CX + 1) whether there are real numbers a, B, C such that the domain of definition is an odd function of R and an increasing function from 1 to positive infinity

Suppose there is such a real number,
Since the function f (x) = log3 (x ^ 2 + ax + b) / (x ^ 2 + CX + 1) is an odd function on R, so
F (0) = log3, B = 0, B = 1
Because the function is an increasing function in [1, + ∞), the,
We discuss the function in this interval separately
f(x)=log3 (x^2+ax+1)/(x^2+cx+1)=log3 【1+(a-c)/(x+1/x+c)】
Since the function is meaningful in R, a ^ 2-4

What is the symmetry axis of the image of the function f (x) = (4 ^ x + 1) and 2 ^ x?

x=0

It is known that the function f (x) = ax ^ 3 + CX (a > 0) obtains the extremum f (x1) at X1 and X2, and the absolute value of x1-x2 is 2, f (x1) - f (x2) = x2-x1
(1) Finding the analytic expression of F (x) (2) finding the monotone interval and extremum of function f (x)

(1) If the function f (x) = ax ^ 3 + CX (a > 0) obtains the extremum at X1 and X2, then f (x1) '= 0, f (x2)' = 0, f (x) '= 3ax ^ 2 + C, then 3A (x1) ^ 2 + C = 0, 3A (x2) ^ 2 + C = O. if the absolute value of 3A (x1 ^ 2-x2 ^ 2) = 0, 3A (x1-x2) (x1 + x2) = 0x1-x2 is 2, a > 0, then X1 + x2 = 0, and | x1-x2 | = 2