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Function y = (1 / 2) ^ √ - x ^ 2 + 2x + 3, domain x belongs to [- 1,3], let t = - x ^ 2 + 2x + 3 Then t belongs to. U = √ T, then u belongs to. Then y = (1 / 2) ^ u belongs to. When x belongs to [1,3], t = - x ^ 2 + 2x + 3 monotone recursion. Y = (1 / 2) ^ u monotone recursion on r.. F (x) monotone recursion It is mainly to explain why "y = (1 / 2) ^ u belongs to" here, y is equal to that. In addition, to find "y = (1 / 2) ^ u belongs to" is to find the function range The questions may be a bit messy, If I get it, I'll do it! U belongs to [0.2], isn't y = (1 / 2) ^ u equal to y = (1 / 2) ^ 0 and y = (1 / 2) ^ 2?
The first space is to find the value range of T, as long as you draw the image of quadratic function, you can calculate that t belongs to [0,4]. The second space, u belongs to [0,2] the third space, and (1 / 2) ^ u belongs to [1 / 16,1] the fourth space, decreasing, and observe the function image. The fifth space, because y = (1 / 2) ^ u is a decreasing function at (0, + ∞), so y increases with the increase of U
On the exponential function of senior one
If the inequality a ^ x2-2ax ＞ (1 / a) ^ x + 1 (a = \ = 1) holds for all real numbers x, then the value range of a is (?)
PS: give the answer, ask for bonus points casually, trumpet, not distressed
a^(x^2-2ax)>a^(-x-1)
a> 1:00
x^2-2ax+x+1>0
(2a-1)^2
In a ^ m / N when a
Exponential function requires base > 0 and not equal to 1
Simply look at this formula, usually first m power and then n power, but there are some problems
For example (- 2) ^ {1 / 2} and (- 2) ^ {2 / 4}
Therefore, generally avoiding such problems is only a matter of agreement, which is meaningless,
The independent variables of exponential function can be irrational numbers
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