Find the domain of the function: $$f(x)= \sqrt{x^2 - 4x - 45}$$
I'm just guessing here; how about if I square everything and then put it in the graphing calculator? Thanks, Lauri
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$\begingroup$Hint: The domain of a function is the set of input values for which the function is defined. Do you know of any values for which the square root is not defined?
$\endgroup$ $\begingroup$The function $f(x)$ is not defined when $$x^2 - 4x - 45 \lt 0,\tag{1}$$ as the square root function is defined in the reals for non-negative reals only. The only valid "input" for the square root function is non-negative real numbers.
If a function $f(x)$ is not defined on an interval(s) = $I \subseteq \mathbb R$, or is not defined for any particular $c\in \mathbb R$, then the interval(s) and/or points at which $f(x)$ is not defined are excluded from the function's domain.
So what are you left with if you exclude all $x\in \mathbb R$ such that $x^2 - 4x - 45 < 0$? This amounts to excluding your solutions to the inequality given in $(1)$. What remains in $\mathbb R$ is then your domain.
Alternatively, your domain consists of all real $x$ such that $$x^2 - 4x - 45 \geq 0\tag{2}$$
Then your task would then be, essentially, to determine the solutions to the equality given by $(2)$. its solution will be your domain.
$\endgroup$ $\begingroup$Note that $x^2-4x-45=(x-2)^2-49$. We completed the square. This is an almost universally useful move when we are dealing with quadratics.
Now when is the thing under the square root sign bad (negative)?
$\endgroup$ $\begingroup$Well, I just happened to run into a mathematician yesterday (in person) and he just pulled the polynomial out of the square root and added the "greater than or equal to"symbol.Because, a square root number can't be a negative number,but it can be zero. Then we just factored it and it factored out to (x+5)(x-9), which leaves the domain at -5 and less and also at 9 and more.I graphed it and it came out to be right. This is the first time I used this site and I don't see where you get the math expressions to type. Maybe because I didn't join as a member? Lauri
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