Find all values $c$ such that $(x+1)/(x^2+2cx+4)$ has domain R

Question

Word for word:

Find all values of $c$ such that $f(x)=\frac{x+1}{x^2+2cx+4}$ has a domain R

---

I really don't know where exactly to start. I'm not sure what it means by "a domain R"

Answer 1

Your function will not be defined if the denominator is zero. You need that: $$x^2 + 2cx + 4 \neq 0$$ for any $x \in \Bbb R$. Which value of $c$ does this? Which value of $c$ makes that polynomial have no real roots? Which value of $c$ makes $\Delta < 0$ there? Now it's on you (;

Answer 2

We have $d(x)=x^2+2cx + 4 = (x+c)^2 +4-c^2$. Since $\lim_{|x|\to \infty} d(x) = \infty$, we must have $d(x) >0$ for all $x$ in order for $f$ to be defined on $\mathbb{R}$ (otherwise $d(x) = 0$ for some $x$, and $f$ would be undefined).

Since the minimum value is, by inspection, $d(-c)=4-c^2$, we require $|c| < 2$.