question 6\nthe per capita consumption of breakfast cereal in the us has the following model that appears…

question 6\nthe per capita consumption of breakfast cereal in the us has the following model that appears above $c(t)=-0.0033t^{3}+0.119t^{2}-0.351t + 11.45$ pounds, where t is the number of years since 1990.\ndecide whether the rate of consumption was changing more rapidly in 2001 or 2007.\n$c(11)=$ \nibs/year\n$c(17)=$ \nibs/year\nthe rate was changing more rapidly in 2007 2001.\nquestion help: message instructor

question 6\nthe per capita consumption of breakfast cereal in the us has the following model that appears above $c(t)=-0.0033t^{3}+0.119t^{2}-0.351t + 11.45$ pounds, where t is the number of years since 1990.\ndecide whether the rate of consumption was changing more rapidly in 2001 or 2007.\n$c(11)=$ \nibs/year\n$c(17)=$ \nibs/year\nthe rate was changing more rapidly in 2007 2001.\nquestion help: message instructor

Answer

Explanation:

Step1: Differentiate the function

Differentiate (C(t)= - 0.0033t^{3}+0.119t^{2}-0.351t + 11.45) using the power rule ((x^n)^\prime=nx^{n - 1}). [ \begin{align*} C^\prime(t)&=-0.0033\times3t^{2}+0.119\times2t-0.351\ &=- 0.0099t^{2}+0.238t - 0.351 \end{align*} ]

Step2: Calculate (C^\prime(11))

Substitute (t = 11) into (C^\prime(t)): [ \begin{align*} C^\prime(11)&=-0.0099\times(11)^{2}+0.238\times11-0.351\ &=-0.0099\times121 + 2.618-0.351\ &=-1.2079+2.618 - 0.351\ &=1.0591 \end{align*} ]

Step3: Calculate (C^\prime(17))

Substitute (t = 17) into (C^\prime(t)): [ \begin{align*} C^\prime(17)&=-0.0099\times(17)^{2}+0.238\times17-0.351\ &=-0.0099\times289+4.046-0.351\ &=-2.8611 + 4.046-0.351\ &=0.8339 \end{align*} ]

Answer:

(C^\prime(11)=1.0591) lbs/year, (C^\prime(17)=0.8339) lbs/year. The rate was changing more rapidly in 2001.