+130518 Sorry...I typed in 14362 instead of 14632 !! My bad.......
Note, that I said that "if" T ≥ 0......I realize that might not be the case.
Yeah....I think you're correct about the "negative" velocity thing...I hadn't considered that. I suppose we would have to know what the graph was intended to show......does the object stop and not move again, or does it stop and reverse course?? (Which would have to be the case if we were considering "negative" velocities.)
Since it's not stated, I'll have to assume the general case and that your answer is the better one........
+130518 This is just a linear function that has a slope of -1504 and a y intercept of 14362
It's domain and range are the same..... (-∞, ∞)
In other words, all "Xs" and "Ys" are covered by this function
P.S. - This looks like a velocity function with a constant deceleration.
If T has to be ≥ 0, then the domain is [0, ≈9.55] - since a "negative" velocity doesn't make sense, and the range would be [0, 14362] - In effect, the object stops after about 9.55 seconds, hours or whatever.
+118724 Is it possible to find the domain and range of V(T)=14632-(1504)T
I just thought i would think about and discuss Chris's answer a little more.
V=14632-1504T If t is time then $$t\ge 0$$
Therefore the biggest V can be is 14632
I know it is 'normal' for only postive times to be considered but I don't know why you have rejected negative velocities.
also note that acceleration is -1504 (I think of this as a backward 'pull') Initially the velocity is positive but it is slowing down at approx t=9.73units of time the objects stops. [that is when 0=14632-1504T]. Where did you get 9.55 from Chris. Is my calculator broken?
$${\mathtt{0}} = {\mathtt{14\,632}}{\mathtt{\,-\,}}{\mathtt{1\,504}}{T} \Rightarrow \left\{ \begin{array}{l}\end{array} \right\}$$ Why didn't this work!
$${\frac{{\mathtt{14\,632}}}{{\mathtt{1\,504}}}} = {\frac{{\mathtt{1\,829}}}{{\mathtt{188}}}} = {\mathtt{9.728\: \!723\: \!404\: \!255\: \!319\: \!1}}$$
It then immediately begins moving backwards (negative velocity) It will then contiue in the negative direction getting faster and faster for ever.
So I think domain $$[0,\inf)$$ Now I have fogotten the LaTex symbol for infinity!
and range $$(-\inf, 14632]$$
That is what I think anyway. Plus, I have never completely understood why time is never considered in a negative direction. Was there no yesterday???
+130518 Sorry...I typed in 14362 instead of 14632 !! My bad.......
Note, that I said that "if" T ≥ 0......I realize that might not be the case.
Yeah....I think you're correct about the "negative" velocity thing...I hadn't considered that. I suppose we would have to know what the graph was intended to show......does the object stop and not move again, or does it stop and reverse course?? (Which would have to be the case if we were considering "negative" velocities.)
Since it's not stated, I'll have to assume the general case and that your answer is the better one........
+118724 Thanks Chris but the first half of your first question is the most correct answer.
We were never told that it was velocity and time!
