Calculating period of a function

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SUMMARY

The function f(t) = 3t, defined on the interval 0≤t≤π, is analyzed for periodicity with the condition f(t) = f(t+1). The discussion concludes that the function is not periodic, as the values contradict the periodic condition when evaluated at specific points. The confusion arises from the interpretation of the interval and the definition of periodicity, leading to the assertion that the period cannot be 1.

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Hi I've been given this question for cooursework and am really struggling, help needed!
heres the question:

What is the period of the function?
f(t) = 3t 0≤t≤π, where f(t)=f(t+1)



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The Attempt at a Solution

 
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What are you struggling with? Do you know what it means for a function to be periodic?
 
yes i understand what it means to be periodic, is the period of the function 1
 
what's the definition of periodicity that you've been given?
 
"What is the period of the function?
f(t) = 3t 0≤t≤π, where f(t)=f(t+1)"

This looks contradictory to me. f(1) = 3, f(2) = 6 contradicts f(t)=f(t+1) when t = 1. You can't have both.
 
3t is only specified as the value of f on the interval [0, n] where n is the period of f.
 
0≤t≤π looks like pi to me from the copy and paste list: π. Doesn't look like 0 ≤ t ≤ n.

Perhaps it was just a typo.
 
Now that you put them next to each other, it does look like pi. I don't think that the function is periodic of period 1 either in that case.
 

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