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November 28, 2014

November 28, 2014

Posted by **Anonymous** on Sunday, April 27, 2008 at 1:48pm.

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

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

Did I do the questions correctly?

- Calculus -
**Damon**, Sunday, April 27, 2008 at 8:09pmaverage = integral over x from x1 to x2 / (x2-x1)

= 5000 int 1.12^x dx from 2 to 3 over 1

int 1.12^x dx = 1.12^x/ln 1.12

ln 1.12 = .1133

1.12^3 = 1.405

1.12^2 = 1.254

(1.405 - 1.254)/.1133 = 1.333

5000 *1.333 = 6663.72

b

f(3) = 5000*1.12^3 = 7024.64

note in part a it is an exponential not a sequence. The value is increasing constantly (presumably) not in quarterly or yearly jumps as in with compound interest in most banks.

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