Answer
time interval [1,4] = $22$
The rate of change at t = 1 is approximately $4$
Work Step by Step
time interval [1,4]
$\frac{ΔS}{Δt}$ = $\frac{s(4)-s(1)}{4-1}$ = $\frac{(4^{3}+4)-(1^{3}+1)}{4-1}$ = $22$
time interval [1,1.01]
$\frac{ΔS}{Δt}$ = $\frac{s(1.01)-s(1)}{1.01-1}$ = $\frac{(1.01^{3}+1.01)-(1^{3}+1)}{1.01-1}$ = $4.0301$
time interval [1,1.001]
$\frac{ΔS}{Δt}$ = $\frac{s(1.001)-s(1)}{1.001-1}$ = $\frac{(1.001^{3}+1.001)-(1^{3}+1)}{1.001-1}$ = $4.0030$
time interval [1,1.0001]
$\frac{ΔS}{Δt}$ = $\frac{s(1.0001)-s(1)}{1.0001-1}$ = $\frac{(1.0001^{3}+1.0001)-(1^{3}+1)}{1.0001-1}$ = $4.0003$
time interval [0.99,1]
$\frac{ΔS}{Δt}$ = $\frac{s(1)-s(0.99)}{1-0.99}$ = $\frac{(1^{3}+1)-(0.99^{3}+0.99)}{1-0.99}$ = $3.9701$
time interval [0.999,1]
$\frac{ΔS}{Δt}$ = $\frac{s(1)-s(0.999)}{1-0.999}$ = $\frac{(1^{3}+1)-(0.999^{3}+0.999)}{1-0.999}$ = $3.9970$
time interval [0.9999,1]
$\frac{ΔS}{Δt}$ = $\frac{s(1)-s(0.9999)}{1-0.9999}$ = $\frac{(1^{3}+1)-(0.9999^{3}+0.9999)}{1-0.999}$ = $3.9997$
The rate of change at t = 1 is approximately $4$
