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Find the definition field (1) f (x) = loga (1 + 3x-4x ^ 2) (a > 1 and a ≠ 1); (2) f (x) = [LG (2x ^ 2 + 5x-12)] / √ (x ^ 2-3)
1、-4x²+3x+1>0
4x²-3x-13
X3 / 2 and X √ 3
So: X √ 3
The domain of definition is (- ∞, - 4) U (√ 3, + ∞)
I wish you happy! I hope I can help you
The function f (x) is a piecewise function f (x) = (1 / 6) * (x ^ 2 + 5x), (0
f(x)=(1/6) *(x^2+5x),( 0
F (x) = - 0.5x Λ 2 + X. whether there is m.n (M is less than n), so that when x belongs to [M, n], the range of function is [2m, 2n]. If there is, m, n can be obtained
No, please explain why. Do me a favor. Isn't it
Existence
F (x) = - x ^ 2 / 2 + x = - 1 / 2 (x-1) ^ 2 + 1 / 2 the axis of symmetry is x = 1;
1,m=
Find the range of function f (x) = x ^ 2-2x-1 / x ^ 2-5x + 6
I reduced this function to f (x) = 1 + 2 / x + 3 + 1 / X-2, but I still can't do it Similarly, the above is also a quadratic trinomial, and the following is also a quadratic trinomial function. How to find the range?
Let me be more standard: I reduced the function to f (x) = 1 + 2 / (x + 3) + 1 / (X-2),
You don't have the right method. It's called delta method
For similar problems, you should multiply the denominator by Y, and then move the numerator to the left to merge the similar items
It is reduced to (Y-1) x ^ 2 - (5y-2) x + 6y + 1 = 0
From the meaning of the question, we can get: (5y-2) ^ 2-4 (Y-1) (6y + 1) ≥ 0
The solution is y ∈ R
If x has a range, let's use the root distribution method. Let's start with the delta method
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