Question

the instantaneous growth rate r of a colony of bacteria t hours after the start of an experiment is given by the function r = 0.01t³ - 0.04t² + 0.01t + 0.06 for 0 ≤ t ≤ 7. find the times for which the instantaneous growth rate is zero.
the instantaneous growth rate is zero at the times □ hour(s).
(use a comma to separate answers.)

Explanation:

Step1: Set the function equal to zero

$0.01t^{3}-0.04t^{2}+0.01t + 0.06=0$
Multiply through by 100 to clear the decimals: $t^{3}-4t^{2}+t + 6 = 0$

Step2: Use the Rational - Root Theorem

The possible rational roots are factors of the constant term 6. The factors of 6 are $\pm1,\pm2,\pm3,\pm6$.
Test $t = - 1$: $(-1)^{3}-4(-1)^{2}+(-1)+6=-1 - 4-1 + 6=0$. So, $(t + 1)$ is a factor.

Step3: Perform polynomial long - division

Divide $t^{3}-4t^{2}+t + 6$ by $t + 1$.
$t^{3}-4t^{2}+t + 6=(t + 1)(t^{2}-5t + 6)$

Step4: Factor the quadratic

Factor $t^{2}-5t + 6=(t - 2)(t - 3)$
So, $t^{3}-4t^{2}+t + 6=(t + 1)(t - 2)(t - 3)=0$

Step5: Solve for t

Set each factor equal to zero:
$t+1 = 0\Rightarrow t=-1$
$t - 2=0\Rightarrow t = 2$
$t - 3=0\Rightarrow t = 3$
But since $0\leq t\leq7$, we discard $t=-1$.

Answer:

$2,3$