Question

question
a researcher tracks a bacterial population in thousands of bacteria over time in hours, and models their rate of change with the function b below:
b(t)=-0.018t^{3}+0.406t^{2}-1.72t + 4.668
b(t) is measured in thousands of bacteria per hour and time t is measured in hours since noon for t on (0,18). based on the model, at what value of t does the rate of change in the bacterial population have a relative minimum? round your answer to 3 decimal places.
answer attempt 1 out of 2

Explanation:

Step1: Find the derivative of $B(t)$

We use the power - rule for differentiation. If $y = ax^n$, then $y'=nax^{n - 1}$.
$B'(t)=\frac{d}{dt}(-0.018t^{3}+0.406t^{2}-1.72t + 4.668)=-0.054t^{2}+0.812t-1.72$

Step2: Find the second - derivative of $B(t)$

Differentiate $B'(t)$ with respect to $t$:
$B''(t)=\frac{d}{dt}(-0.054t^{2}+0.812t - 1.72)=-0.108t + 0.812$

Step3: Find the critical points of $B'(t)$

Set $B'(t) = 0$. We use the quadratic formula for $ax^{2}+bx + c = 0$, which is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a=-0.054$, $b = 0.812$, and $c=-1.72$.
$t=\frac{-0.812\pm\sqrt{(0.812)^{2}-4\times(-0.054)\times(-1.72)}}{2\times(-0.054)}$
$t=\frac{-0.812\pm\sqrt{0.659344 - 0.37344}}{-0.108}$
$t=\frac{-0.812\pm\sqrt{0.285904}}{-0.108}$
$t=\frac{-0.812\pm0.5347}{-0.108}$
We get two solutions:
$t_1=\frac{-0.812 + 0.5347}{-0.108}=\frac{-0.2773}{-0.108}\approx2.568$
$t_2=\frac{-0.812-0.5347}{-0.108}=\frac{-1.3467}{-0.108}\approx12.470$

Step4: Use the second - derivative test

Evaluate $B''(t)$ at the critical points.
For $t = 2.568$:
$B''(2.568)=-0.108\times2.568+0.812=-0.277344 + 0.812 = 0.534656>0$
For $t = 12.470$:
$B''(12.470)=-0.108\times12.470+0.812=-1.34676+0.812=-0.53476<0$
Since $B''(t)>0$ at $t = 2.568$, the function $B'(t)$ has a relative minimum at $t\approx2.568$.

Answer:

$2.568$