Is there a faster way to find a range of a specific function?

[abel muroi]

For example

If i am given a simple function like... f(x) = SquareRoot(x + 4)

I usually just graph the function and find the range from there. Which can sometimes be a bit tedious when I am in the middle of a test.

But is there a method that i can use to find the range WITHOUT graphing it?

 

[pka]

For example
If i am given a simple function like... f(x) = SquareRoot(x + 4)
I usually just graph the function and find the range from there. Which can sometimes be a bit tedious when I am in the middle of a test.
But is there a method that i can use to find the range WITHOUT graphing it?

Yes the is. But it takes a ton of experience; doing hundreds of these problems.

Consider \(\displaystyle S(x)=\sqrt{f(x)}\) then the \(\displaystyle Dom(S)=\{x~:~f(x)\ge 0\}\).
If \(\displaystyle \displaystyle{\lim _{x \to \infty }}f(x) = \infty \) then \(\displaystyle \displaystyle{\lim _{x \to \infty }}Rng(S) = \infty \)

 

[ksdhart]

My process for finding the range of a function is to start at the very beginning. You, of course, know that the range of a function is set of all possible outputs of said function. The three main things to consider are any horizontal asymptotes of the function, and the behavior as x approaches infinity, and as x approaches negative infinity. For some functions, it might be helpful to look at what happens as x approaches 0 from either side, as well. Over time, you'll begin to note certain tips and tricks as well to save yourself even more time. I'll demonstrate two that I know of, using example functions:

\(\displaystyle y=sqrt{x^3+8}\)

The first thing I note is that anything under a radical (of any power: square roots, cube roots, 17th roots, etc.) can never be negative, so that means I can instantly say that y >= 0. Because y can't be negative, I need only look at its behavior as x approaches infinity. If x could ever get to infinity, the function would equal the square root of infinity, or still infinity. In other words, as x grows larger, so too does y. Thus, the range of the function is all positive real numbers and 0, or \(\displaystyle y\in\left[0,\infty\right)\).

\(\displaystyle y=-370x^{18}-786x^{16}-1235x^{12}-7x^6-27x^2\)

This function looks like an incredibly daunting problem and one that would take ages to determine the range of. However, with one simple trick I can solve this in a flash. This is an even function with all negative coefficients, so it can never be positive (conversely, an even function with all positive coefficients can never be negative). Then I note that for any even positive power of n, \(\displaystyle x^n=(-x)^n\). Thus, the behavior of the function as x approaches negative infinity is the same as the behavior as x approaches positive infinity. So, I need to consider what happens as x grows infinitely large. As before, I consider what would happen if x could actually reach infinity. I see that as x grows larger and larger, the x¹⁸ term dominates the function, and the other terms matter less and less. And thus, ultimately, I'm left with \(\displaystyle -370x^{\infty }=-\infty\). So, the range is then \(\displaystyle y\in \left(-\infty ,0\right]\)