pka
Elite Member
- Joined
- Jan 29, 2005
- Messages
- 11,971
Well I completely disagree that one needs to have a graph to work this question.
[imath]w(x)=\begin{cases}-x^3+1 & x\le 1 \\2|x-1|-2 & 1<x<3\\ (x-5)^2-2 & 3\le x \end{cases}[/imath]
Look the right column. We have [imath](-\infty, 1]\cup(1,3)\cup [3,\infty)=\Re[/imath], the set of real numbers.
That tells us that the set of all real numbers is the domain of [imath]w(x).[/imath]
Now we have:
[imath]1)~\mathop {\lim }\limits_{x \to - \infty } w(x) = +\infty [/imath]
[imath]2)~\mathop {\lim }\limits_{x \to 1^{{\large\bf-}} } w(x) = 0[/imath]
[imath]3)~\mathop {\lim }\limits_{x \to 1^{{\large\bf+}} } w(x) = -2 [/imath]
[imath]4)~\mathop {\lim }\limits_{x \to 3^{{\large\bf-}} } w(x) = 2 [/imath]
[imath]5)~\mathop {\lim }\limits_{x \to 3^{{\large\bf+}} } w(x) =2 [/imath]
[imath]6)~\mathop {\lim }\limits_{x \to \infty } w(x) = +\infty [/imath]?
Looking at the above limits, tell us what the range is.
