Then we can make an analogy.
1. Use the inverse function method
Application type: molecule. Functions with only one denominator (that is, the rational fraction of one term) can also be used for other function types that are easy to solve the independent variables in reverse.
(The range of the original function is determined by the domain of the inverse function)
Solution: It is easy to know that the domain of the original function is:
It is easy to get the inverse function expression of the original function: x = (3y+5)/(5y-3).
By x>- 1 and x≠3/5
The following set of inequalities can be obtained:
(3y+5)/(5y-3)>- 1
(3y+5)/(5y-3) ≠3/5
solve
Y & gt3/5 or y <-1/4.
therefore
The range of the original function is y∈(-∞,-1/4)U(3/5, +∞).
Combine numbers and shapes.
y = |x+5|+|x-6|
The above formula can be regarded as the sum of the distances from point P(x) to fixed points A(-5) and B(6) on the number axis.
As can be seen from the figure, when the point P is on the line segment AB,
y = | x+5 |+| x-6 | = | AB | = 1 1
When point P is on the extension line or the reverse extension line of AB,
y = |x+5|+|x-6| >|AB| = 1 1
So the function value domain is y∈[ 1 1, +∞).
3. Combine direct observation with numbers and shapes (and exchange elements)
Let t =-x 2+x 2.
Then t =-(x- 1/2) 2+9/4 and t≥0.
Draw the graph of the function t(x). According to the nature of parabola, there are
0≤t≤9/4
Because y = 4-√t is a monotone function.
Then the original function value domain is y ∈ [5/2,4]
Use alternative methods
Let t =√( 1-2x) ≥0, then x = 1/2- 1/2t 2.
The original function is converted to
y = 1/2 - 1/2t^2+t = - 1/2(t- 1)^2+ 1
It is easy to know by graphics.
y≤ 1
So the function range is {y | y ≤ 1}.