Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Like, is their an algebraic method? For example if I am asked to find the domain of $g(t) = \sqrt{t^2 + 6t}$ , how do I determine the range of this?

Is their a universal algebraic method that I don't know about?

share|cite|improve this question
up vote 0 down vote accepted

A general method would be this:
Let $$y=\sqrt{t^2+6t}\\y^2=t^2+6t \\ t^2+6t-y^2=0\\t=\frac{-6\pm\sqrt{36+4y^2}}{2}$$
For t to be real, $36+4y^2\ge0\implies y\in \mathbb{R}$

But $y\not\lt0$, since $y$ is equal to the square root of a real number.
So the range of the function will be $[0,\infty)$.

A more specific method for $f(x)=\sqrt{ax^2+bx+c}$ :
First of all, range($R_f$) $\subseteq [0,\infty)$.
The range of $ax^2+bx+c$ is $[-\frac{b^2-4ac}{4a},\infty)$ if $a>0$ and $(-\infty,-\frac{b^2-4ac}{4a}]$ if $a<0$.
So, the range of $f(x)$ will be the square root of bounds of intersection of $[0,\infty)$ and the range for $ax^2+bx+c$.

Here are some of the "common rules" for $f(x)$ to be real:
1. If $f(x) = \frac 1a$, $a\ne 0$.
2. If $f(x)=\sqrt{a}, a\ge0$.
3. If $f(x)=\frac1{\sqrt{a}}, a>0$
4. If $f(x) = \log_yx, x>0,y>0,y\ne1$ (Didn't want to add this rule since it is very specific)
To find domain of a function, $f(x)$, find for what values of $x$, $f(x)$ will be undefined/not real. To find range, the general method is to find $x$ in terms of $f(x)$ and then find values of $f(x)$ for which $x$ is not defined.

share|cite|improve this answer
Dude, I love you so much. You may have potentially saved me hours of frustration. Thank you so much! I will name my first born child in honor of you (that part was definitely a lie). – Ishamel May 28 '14 at 12:01
@Ishamel I'm glad, I was of some help to you(definitely not a lie). – shaurya gupta May 28 '14 at 13:14

Assuming that you are looking at a real function of a real variable you can determine the allowed domain as those values of t that produce a real result for g. In this case you need $t^2 + 6t \ge 0$ otherwise you are trying to get the square root of a negative number. Factorising $t(t+6) \ge 0$ with solutions $t \ge 0$ and $t \le -6$. So the allowed domain is $t \le -6$ and $t \ge 0$. The corresponding range is the values that g ranges over given this domain which can be seen to start from 0 (if $t = 0$ or $t=-6)$ and extend to $+\infty$

This changes if you allow g to be a complex function of a real variable, or a complex function of a complex variable.

The domain of a function is also often specified as a subset of the allowed domain, so you might have a function like g restriced by definition to a range $t \ge 0$.

share|cite|improve this answer
I think I may have actually understood something, so you basically factored it out to t(t + 6) => 0 and solved for t? But why did you use the => sign? Do you ALWAYS have to use this sign, or only when the original quesiton has a square root in it? – Ishamel May 28 '14 at 10:44
@Ishamel Yes, because for $y=\sqrt{x}, x$ must be $=>0$, since we want $y$ to be real. – shaurya gupta May 28 '14 at 11:00
@shaurya gupta I kind of get it thanks, Is their a general collection of rules such as the one you just mentioned for example in y = square root x the rule is that square roots have to be positive (excluding imaginary numbers..). I have a weak mathematical foundation, and it's those 'tiny' bits of information that hold me back every single time. Anyone is welcome to answer this because it would solve all my maths dilemmas. – Ishamel May 28 '14 at 11:05
@ishamel Let me post an answer. – shaurya gupta May 28 '14 at 11:06
The allowed domain depends on the function. For example, looking at arcsin (i.e. $sin^{-1}$) as a real function of a real variable, then $arcsin(t)$ is only valid for $-1 \le t \le +1$. $log(t)$ is only valid for $t \gt 0$ – Tom Collinge May 28 '14 at 11:09

$t^2 + 6t \geq 0$. Thus $g(t) \geq 0$. This gives the range $[0, +\infty)$. For more "details" about the range, take a non-negative real number $r \geq 0$, then show that:

you can find an $t$ with $t \leq -6$ or $t \geq 0$ such that: $g(t) = r$. This translates to the equation:

$\sqrt{t^2 + 6t} = r \Rightarrow t^2 + 6t = r^2 \Rightarrow (t + 3)^2 - 9 = r^2 \Rightarrow (t + 3)^2 = r^2 + 9 \Rightarrow t + 3 = \pm \sqrt{r^2 + 9} \Rightarrow t = - 3 \pm \sqrt{r^2 + 9}$. Either value of $t$ just found works !

share|cite|improve this answer

Writing $g(t)=\sqrt{t(t+6)}$, you can see that the domain must be restricted to values of $t$ for which $t(t+6)\geq 0$. You can solve this inequality to see that the domain then is $(-\infty,-6]\cup[0,\infty)$. This just excludes the open interval $(-6,0)$ on which $t(t+6)<0$.

So what is the image of $g$? (Note: the range can be taken to be $(-\infty,\infty)$, but I think you want only the more restricted set of values actually assumed by $g$; the terminology is not standardized, unfortunately--but I know what you mean.) Certainly $g(0)=0$, and all values of $g$ are in $[0,\infty)$ because the square root function always returns a nonnegative value. But does $g$ assume all values in $[0,\infty)$? The answer is "yes". You will have to accept that $g(t)$ can be made as large as you like by taking $t$ sufficiently large, and there are no "holes" in the image because $g((-\infty,-6])$ meets $g([0,\infty))$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.