Question

Suppose f(t)=t+1 for -1 13 years ago

Answer 1

...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<a<1,then solve for t,and t is just 2<t<4

Answer 2

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?!!

Answer 3

i do not see the domain changing it is just the variable changing from t TO t-3

Answer 4

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.

Answer 5

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.

Answer 6

Edit: *f(u) has a domain of 2 to 4.

Answer 7

i stick with basking

Answer 8

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.

Answer 9

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:



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.