anonymous
  • anonymous
Suppose f(t)=t+1 for -1
MIT 18.01 Single Variable Calculus (OCW)
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
...That is just a high school year one`s question... I think you are eight,though I am not good at it.Just set t-3=a,where -1
anonymous
  • anonymous
Xavi! Thx for your reply! But you know what? I'm a hundred percent sure that the domain would change and what I (and you) wrote as an answer is right! Actually I post this question to see what others might possibly think of that! This question came up with my mind when I was reading a calculus book by a math prof of MIT and surprisingly saw that the author has considered both functions to have the same domain!!!! What I said then was: "What the heck?!! Duh!" Why should a math prof of MIT say such a thing in his book?!!
anonymous
  • anonymous
i do not see the domain changing it is just the variable changing from t TO t-3

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anonymous
  • anonymous
But I do see! Domain is the input, so now it's not only variable but also the domain which has changed. f(t) has inputs between 1 and -1, meanwhile f(t-3) has inputs t-3 that obviously is a change in inputs, that is, domain! So, we shouldn't look at only symbols but the concepts behind them.
anonymous
  • anonymous
Not quite. If you were to define a new variable u=t-3, you would say that u has a domain of 2 to 4. But the domain over which t varies is the same. It's a matter of how rigorous you want to be and in what way you want to set up your functions. It could be said that f(t) = t+1, g(t) = t-3. The domain of (f(g(t)) is still -1 to 1, because it is the domain of g.
anonymous
  • anonymous
Edit: *f(u) has a domain of 2 to 4.
anonymous
  • anonymous
i stick with basking
anonymous
  • anonymous
In case clarification is necessary: if you had a function f(t) = t+1 and considered f(t-3) defined on [-1,1], you would be talking about the line that extended from the point (-1,-3) [f(-1-3) = -4+1 = -3] to the point (1,-1) [f(1-3) = -2+1 = -1]. In other words, the input isn't (t-3); the input is still t.
anonymous
  • anonymous
basking and pasta! I'll reason based on your reasoning line. Now, put it this way: 1. When we say f(t) = t + 1 for -1 < t <1, it means the function is defined only on that interval. In other words, the domain or INPUT of f(t) must belong to (-1,1), no matter what symbol you use! The input could be either just a variable like t or z OR a function like g(t) or h(z). Whatever they were, they'd have to belong to the interval (-1,1). (RIGHT?) 2. I suppose you nodded! We replace the input of f(t) with g(t). g(t) itself is a function that is defined to be t-3, i.e. g(t) = t - 3. Thus now, our function is like f(g(t)) with g(t) as its input, which according to No. (1) (the 1st paragraph) has to be between -1 and 1 or -1 < g(t) < 1. 3. Thanks to g(t) = t - 3, we have -1 < t - 3 < 1 that implies 2 < t < 4!! It does make sense! Because we changed the input of f(t) so naturally the domain in terms of t, changes. But a graph says a thousand words!! So I'd like to show geometrically that how the domain changes. Look at the graph below: |dw:1350812103357:dw| The function just moved 3 units to the right that clearly the domain changed to keep the input of the function between -1 and 1.
anonymous
  • anonymous
Sorry for the graph! Don't know why the graph is out of screen! Anyway, to clarify the unseen parts, the horizontal axis is 't' and the right line is the graph of 'f(t-3).
anonymous
  • anonymous
Here's the confusion: Is t limited or is the domain of f limited? These are not the same thing. The problem statement is ambiguous. Example 1: If t is the sea level, and it's between -1 and 1 meter and we consider the domain of f(t) = t+1 and f(t-3), then they'd both be (-1,1), unless you considered the domain of f(t-3) to be (-4,-2). Example 2: f(x)=arcsin(x). The domain of f(x-3) is the set of x-values [2,4].
Stacey
  • Stacey
Basking is correct with those two examples. The domain depends on the problem. I would tend to follow example 2 in most cases and figure that the domain is limited by the function (trig, denominator can't be zero, or some other limiting factor).
anonymous
  • anonymous
Good point, basking! And Stacey! Thanks for your reply! I considered the domain of f(t) to be limited when posted my question! And I guess that's the first thing that comes in everybody's mind when seeing such question in math! Otherwise, I'd mention that. I took a glance at that book again, and found an example of that section concerning such topics. I'll attach a shot of a related graph from the book to this post. But about Example 1, I ought to say that that's a practical situation, while we're talking about a problem in pure math. In my book, the Example 2 works in here. That's what I'm talking about.
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anonymous
  • anonymous
i had ask my tutor this question(does the domain cahnge).he said YES the variable and the domain change but the two equations are equal.
anonymous
  • anonymous
Thank you very much pasta for doing so! And I'm happy with her answer :) But what do you (or does she) mean by saying the two equations are equal?! Could you clarify that? Notice that here we have functions NOT equations. In math, we say two functions f(x) and g(x) are equal if and only if they meet two conditions below: 1. \[D_{f} = D_{g}\]2. \[f(x) = g(x)\] So?!
anonymous
  • anonymous
the two functions are the same
anonymous
  • anonymous
pasta! If the domain change (as your teacher said), how could it be all possible for two functions to be the same?!! If domain changes, it means it's a different function, according to the conditions I mentioned earlier.
anonymous
  • anonymous
Again we have ambiguity. f does not change. But f(t-3) is not the same function of t as f(t).
anonymous
  • anonymous
THE FUNCTIONS BEHAVE SIMILLARLY
anonymous
  • anonymous
Maybe it's too late to say that, but I'd like to thank you all for taking your time and getting involved in this discussion :) My Final Conclusion: The domain of the function does change! And f(t) is not the same as f(t-3); In mathematical terms, these two functions are NOT equal. But (as pasta exquisitely mentioned) the functions behave similarly!

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