Answer with Step-by-step explanation:
We are given that
Initial deposit=$100
Interest rat=r=7%
[tex]f(t)=100(1.07)^t[/tex] (in dollars)
Where t(in years)
a.We have to find the rate of change of f(t).
f(t) varies with t Where f(t) in years and t in years
Therefore, the unit of rate of change of f(t) is dollar/year.
b.We have to find the average rate of change over [0,-0.5] and [0,1].
Substitute t=0
[tex]f(0)=100(1.07)^0=100[/tex]
Substitute t=0.5
[tex]f(0.5)=100(1.07)^{0.5}=103.44[/tex]
Substitute t=1
[tex]f(1)=100(1.07)=107[/tex]
Average rate on interval [a,b] =[tex]\frac{f(b)-f(a)}{b-a}[/tex]
Using the formula
Average rate on interval [0,0.5]=[tex]\frac{103.44-100}{0.5-0}[/tex]
Average rate on interval [0,0.5]=[tex]\frac{3.44}{0.5}=6.88[/tex]dollar/year
Average rate on interval [0,1]=[tex]\frac{107-100}{1-0}=7dollar/year[/tex]
c.Average rate on interval [0.5,1]=[tex]\frac{107-103.44}{1-0.5}=7.12[/tex]dollars/year
Instantaneous rate of change at t=0.5 =Average of the average values obtained on interval [0,0.5] and [0.5,1]=[tex]\frac{6.88+7.12}{2}=7dollar/year[/tex]
