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?
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$\begingroup$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. $\endgroup$ 2 $\begingroup$
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 $t \ge 0$.
$\endgroup$ 5 $\begingroup$$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 !
$\endgroup$ $\begingroup$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 $g([0,\infty))$.
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