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1. If the domain of definition of function f (x) = quadratic radical (AX + 2) is (- ∞, 1], find the real number a 2. If the function f (x) = quadratic root sign (AX + 2) is significant in (- ∞, 1], find the real number a 3. If the domain of definition of function f (x) = quadratic root (AX & sup2; + BX + C) is [1,3], find the domain of definition of function f (x) = quadratic root (BX & sup2; - ax + C) I don't understand very well. It's better to have explanation and steps,
1. When x = 1, ax + 2 is exactly equal to 0, so a = - 2
2. It is shown that ax + 2 is always greater than or equal to 0 in the (- ∞, 1] interval, so a is less than or equal to - 2
3. From the meaning of the title, we can know that the condition that ax & sup2; + BX + C is always greater than or equal to 0 is that x indicates that a is less than 0 and ax & sup2; + BX + C = 0 is 1 and 3 in [1,3] interval, so - B / a = 1 + 3 = 4, C / a = 1 * 3 = 3 (did you learn Weida's theorem?) so B is less than 0
The following is to find the value that BX & sup2; - ax + C is greater than or equal to 0, because B is less than 0, so the sum of the two is = - (- a) - B = - 1 / 4, and the product of the two is C / b = (C / a) \ (B / a) = 3-4 = - 3 / 4
So two are The answer is too complicated. I believe you can work it out namely
The problem of finding two numbers Mn = - 3,4 m + n = - 1,4
Let one of the unknowns be replaced by another to solve a quadratic equation of one variable
Finally, x ^ 2 + x-3 = 0
Do it for yourself
You can make do with our hard work
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