The solution of high school function range

The solution of high school function range

1. The observation method is used for simple analytic expressions. Y = 1 - √ x ≤ 1, range (- ∞, 1] y = (1 + x) / (1-x) = 2 / (1-x) - 1 ≠ - 1, range (- ∞, - 1) ∪ (- 1, + ∞). 2. The collocation method is mostly used for quadratic (type) functions. Y = x ^ 2-4x + 3 = (X-2) ^ 2-1 ≥ - 1, range [- 1, + ∞) y = e ^ 2x-4e ^ x-3 = (e ^ X-2) ^ 2-7 ≥ - 7

High school function range
Find the range of (x) + (1-x) under y = root sign

1. Trigonometric substitution
The domain is [0,1]
Let x = (Sina) ^ 2
y=sina+cosa
According to the auxiliary angle formula, y = root 2Sin (a + 45 degrees), a belongs to [0,90]
So 1

High school function range
In order to find the range of a function in the form of y = (AX2 + BX + C) / (MX2 + NX + P) (at least one of a and M is not zero), we can use the discriminant method to find the range, but we should pay attention to the following problems: check whether the equation has a solution when the coefficient of quadratic term is zero, if there is no solution or the function is meaningless, We should remove the value from the range; the boundary of the closed interval should also consider the existence of X when it reaches the value; the numerator denominator must be a reduced fraction
When the coefficient of quadratic term is zero, how can the equation have no solution or make the function meaningless? Please give an example;
Why should we consider the existence of X for the boundary of closed interval? Please give an example;
Why can't I use the irreducible fraction

If your topic is for any function, and the domain is a natural domain, that is, X can take all real numbers, as long as the denominator of the fraction is not zero, then the method is as follows: because the denominator of the fraction is not zero, first transform the equation, MYX ^ 2 + NYX + py = ax ^ 2 + BX + C, shift and merge: (my