Here's a method without calculus.
Let 
Denominator can never be 
in this case. So the domain depends only upon $4y^2-24+4y\Rightarrow 4y^2+4y-24≥0\Rightarrow 4(y-2)(y+3)≥0\Rightarrow (y-2)(y+3)≥0\Rightarrow y\in(- \infty, -3]\cup[2,\infty).$
Hence the range of the 
is
I guess you know solving inequalities...
Answer from An Alien on Stack ExchangeHere's a method without calculus.
Let
Denominator can never be in this case. So the domain depends only upon $4y^2-24+4y\Rightarrow 4y^2+4y-24≥0\Rightarrow 4(y-2)(y+3)≥0\Rightarrow (y-2)(y+3)≥0\Rightarrow y\in(- \infty, -3]\cup[2,\infty).$
Hence the range of the is
I guess you know solving inequalities...
To find the range of a function, my first instinct is to check whether the graph has an inverse. We can do this informally using the Horizontal Line Test. If no Horizontal Line intersects the function more than once, then the function has an inverse. With that, we can find 
. The "domain" of 
would be the range of the function 
.
Else, we can use the derivatives of the function to help us find the maximum and minimum which gives us the lower and upper bound of the range of 
.
Hey so, I jumped a math class and I have a big test coming. I know how to find the domain of a function without a graph, but how do you find the range? The range of a function without a graph, I mean. They discussed it in the class before this one but I wasn't there and the notes aren't telling me much. If there isn't a specific way, are there any tips or tricks? Specifically quadratic functions or functions with square roots in them? My teacher in his notes just replaces the radicand or any other number in a parentheses like thing with [0, infinity)
How can you find the range of a function without just simply observing it's graph, are there any analytical methods?
f(x) = x-2, x ≤ 3
Then,
f(x) ≤ f(3)
f(x) ≤ 3-2
f(x) ≤ 1
It leads to the correct answer.
2. g(x) = 2x², -3 ≤ x ≤ 1
Then,
g(-3) ≤ g(x) ≤ g(1)
2(-3)² ≤ g(x) ≤ 2(1)²
18 ≤ g(x) ≤ 2
In this case, of course, it makes little sense (at least to me). Of course, by drawing a graph, you easily get the correct answer, 0 ≤ g(x) ≤ 18.
I have two questions:
Why does this not work? What's really going on?
Is it possible to do without drawing a graph? How?
The function $f(x) = 1/x$ has domain $D_f = (-\infty, 0) \cup (0, \infty)$ and range $R_f = (-\infty, 0) \cup (0, \infty)$.
Based on your work, we can see that the graph of the function $$g(x) = \frac{x - 1}{x + 2} = 1 - \frac{3}{x + 2}$$ is obtained from the graph of $f(x)$ by reflecting the graph of $f(x)$ in the $x$-axis, stretching it vertically by a factor of $3$, and translating it two units to the left and one unit up. Consequently, the domain of $g(x)$ is $D_g = (-\infty, -2) \cup (-2, \infty)$ and its range is $R_g = (-\infty, 1) \cup (1, \infty)$.
As an appendix to Nikolaos Skout's hint, the function looks like below. You should be able to imagine something like this using the usual methods and tricks in calculus for graphing functions.