How Do You Calculate the Average Rate of Change for the Function g(t)?

AI Thread Summary
To calculate the average rate of change (ARoC) for the function g(t) = 1/(3t-2) between t = 0 and t = a + 1, the formula ARoC = (f(b) - f(a)) / (b - a) is applied. The values f(0) and f(a + 1) are computed as -1/2 and 1/(3a + 1), respectively. The difference f(a + 1) - f(0) is simplified to (1/(3a + 1) + 1/2), and a common denominator of 6a + 2 is used to combine the fractions. The final expression simplifies to 3/(6a + 2), representing the average rate of change. The calculations confirm that the approach to finding the ARoC was correct.
willywonka12345
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Homework Statement



For the function g(t) = 1/(3t-2) determine the average rate of change between the values t = 0 and t = a + 1


Homework Equations



ARoC formula : f(b) - f(a) over b-a

The Attempt at a Solution



So I think I am doing it right but can only get so far and then get stuck.

I set it up : 1/3(a+1)-2 (-) 1/3(0)-2 all over a+1-0
I calculate and get : 1/3a+1 (+) 1/2 all over a+1
I get a common denominator to add 1/3a+1 (+) 1/2 : 2/6a+2 (+) 3a+1/6a+2 all over a+1
and then I am not sure, or even what I have is close to coming correct. Please help. :smile:
 
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I would prefer you used parentheses to clarify! f(0)= -1/2 certainly and
f(a+1)= 1/(3(a+1)-2)= 1/(3a+1) so f(a+1)- f(0)= 1/(3a+1)+ 1/2. As you say, the common denominator is 2(3a+1)= 6a+ 2.

2/(6a+2)+ (3a+1)/(6a+2)= (3a+3)/(6a+2)= 3(a+1)/(6a+2). It's easy to divide that by a+1.
 
HallsofIvy said:
I would prefer you used parentheses to clarify! f(0)= -1/2 certainly and
f(a+1)= 1/(3(a+1)-2)= 1/(3a+1) so f(a+1)- f(0)= 1/(3a+1)+ 1/2. As you say, the common denominator is 2(3a+1)= 6a+ 2.

2/(6a+2)+ (3a+1)/(6a+2)= (3a+3)/(6a+2)= 3(a+1)/(6a+2). It's easy to divide that by a+1.

:redface: Sorry, Ok, I made it harder than it was I think, for the final answer you get : 3/(6a+2) :) ??
 
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