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) = -0.031t^3 + 1.095t^2 - 11.191t + 40.578 \\)
\\( b(t) \\) is measured in thousands of bacteria per hour and time \\( t \\) is measured in hours since 6 a.m. for \\( t \\) on \\( 0, 22 \\). based on the model, at what value of \\( t \\) does the rate of change in the bacterial population change signs? round your answer to 3 decimal places.

Explanation:

Step1: Find the roots of \(B(t) = 0\)

We have the cubic equation \( - 0.031t^{3}+1.095t^{2}-11.191t + 40.578=0\). We can use the Newton - Raphson method. The Newton - Raphson formula is \(t_{n + 1}=t_{n}-\frac{B(t_{n})}{B^{\prime}(t_{n})}\). First, find the derivative \(B^{\prime}(t)=-0.093t^{2}+2.19t - 11.191\).
Let's start with an initial guess. We can also use a graphing utility or a calculator with a root - finding function.

Step2: Calculate the roots

Using a calculator or software (such as a TI - 84 Plus with the zero function or a computer algebra system like Mathematica with NSolve), we find the roots of the equation \( - 0.031t^{3}+1.095t^{2}-11.191t + 40.578 = 0\).
The roots of the cubic equation \(ax^{3}+bx^{2}+cx + d=0\) (here \(a=-0.031\), \(b = 1.095\), \(c=-11.191\), \(d = 40.578\)) are found as follows:

$$t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$$

(for the quadratic part after factoring out a root or using a root - finding algorithm).
After calculation, we get \(t\approx5.999\)

Answer:

\(t\approx6.000\)