average = 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.
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
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?
1 answer