Answer
a) $dollars/year$
b)
Find the average rate of change over
[0, 0.5] = $7.8461$
[0, 1] = $8$
c) The rate of change at t = 0.5 is approximately $8/yr
Work Step by Step
a) $\frac{Δf}{Δt}$ = $\frac{dollars}{year}$ = $dollars/year$
b)
[0, 0.5]
$\frac{Δf}{Δt}$ = $\frac{100(1.08)^{0.5}-100(1.08)^{0}}{0.5-0}$ = $7.8461$
[0, 1]
$\frac{Δf}{Δt}$ = $\frac{100(1.08)^{1}-100(1.08)^{0}}{1-0}$ = $8$
c)
time interval [0.5, 0.51]
$\frac{Δf}{Δt}$ = $\frac{100(1.08)^{0.51}-100(1.08)^{0.5}}{0.51-0.5}$ = $8.0011$
time interval [0.5, 0.501]
$\frac{Δf}{Δt}$ = $\frac{100(1.08)^{0.501}-100(1.08)^{0.5}}{0.501-0.5}$ = $7.9983$
time interval [0.5, 0.5001]
$\frac{Δf}{Δt}$ = $\frac{100(1.08)^{0.5001}-100(1.08)^{0.5}}{0.5001-0.5}$ = $7.9981$
time interval [0.49, 0.5]
$\frac{Δf}{Δt}$ = $\frac{100(1.08)^{0.5}-100(1.08)^{0.49}}{0.5-0.49}$ = $7.9949$
time interval [0.499, 0.5]
$\frac{Δf}{Δt}$ = $\frac{100(1.08)^{0.5}-100(1.08)^{0.499}}{0.5-0.499}$ = $7.9977$
time interval [0.4999, 0.5]
$\frac{Δf}{Δt}$ = $\frac{100(1.08)^{0.5}-100(1.08)^{0.4999}}{0.5-0.4999}$ = $7.998$
The rate of change at t = 0.5 is approximately $8/yr
