# Limits - Did I do the questions correctly?

If you invest $5000 in a stock that is increasing in value at the rate of 12% per year, then the value of your stock is given by: f(x) = 5000(1.12)^x, where x is measured in years a) find average value from x = 2 to x = 3 b) find instantaneous value at x = 3 ## Homework Equations y = b^x y' = (b^x)(ln^x) ## The Attempt at a Solution a) f(x) = 5000(1.12)^{5 - 2} = 5000(1.12)^3 = 7024.64 b) f'(x) = (5000(1.12)^3)(ln 1.12) = (7024.64)(0.113) = 796.093 ## Answers and Replies HallsofIvy Science Advisor Homework Helper ## Homework Statement If you invest$5000 in a stock that is increasing in value at the rate of 12% per year, then the value of your stock is given by:

f(x) = 5000(1.12)^x, where x is measured in years

a) find average value from x = 2 to x = 3
b) find instantaneous value at x = 3

## Homework Equations

y = b^x
y' = (b^x)(ln^x)
No. the derivative of b^x is ln(b) b^x. I guess that was a typo.

## The Attempt at a Solution

a) f(x) = 5000(1.12)^{5 - 2} = 5000(1.12)^3 = 7024.64
b) f'(x) = (5000(1.12)^3)(ln 1.12) = (7024.64)(0.113) = 796.093[/QUOTE]
I have no idea why you did either of those things. You are asked for the "average value" of f(x) between x= 2 and x= 3 and you calculated f(3)? In what sense is that an average? What is the standard definition of "average" of a continuous function? Is it at all likely that the "average" value of an incerasing function will be the very last value?

For (b), why did you differentiate? there is nothing said obout a derivative, only the "instantaneous" value of the function. Do you really think the value at x= 3 is lower than the amount you started with?

HallsofIvy
Homework Helper
No. the derivative of b^x is ln(b) b^x. I guess that was a typo.

b) f'(x) = (5000(1.12)^3)(ln 1.12) = (7024.64)(0.113) = 796.093
I have no idea why you did either of those things. You are asked for the "average value" of f(x) between x= 2 and x= 3 and you calculated f(3)? In what sense is that an average? What is the standard definition of "average" of a continuous function? Is it at all likely that the "average" value of an incerasing function will be the very last value?

For (b), why did you differentiate? there is nothing said obout a derivative, only the "instantaneous" value of the function. Do you really think the value at x= 3 is lower than the amount you started with?

You may be completely misunderstanding the question- I cannot see that it has anything to do with "limits".

I thought instantaneous value means the same thing as slope and slope means the same thing as finding the derivative?

My teacher wanted the class to use those formulas mentioned and yes it was a typo..it's y' = (b^x)(lnb)