the number of visitors ( p ) to a website in a given week over a 1 - year period is given by ( p(t)=121+(t…

the number of visitors ( p ) to a website in a given week over a 1 - year period is given by ( p(t)=121+(t - 88)e^{0.02t} ), where ( t ) is the week and ( 1leq tleq52 ).\na) over what interval of time during the 1 - year period is the number of visitors decreasing?\nb) over what interval of time during the 1 - year period is the number of visitors increasing?\nc) find the critical point, and interpret its meaning.\na) the number of visitors is decreasing over the interval ( (1,38) ).\n(simplify your answer. type integers or decimals rounded to three decimal places as needed. type your answer in interval notation.)\nb) the number of visitors is increasing over the interval ( (38,52) ).\n(simplify your answer. type integers or decimals rounded to three decimal places as needed. type your answer in interval notation.)\nc) the critical point is ( (38,14.086) ).\n(type an ordered pair. type integers or decimals rounded to three decimal places as needed.)\ninterpret what the critical point means.\nthe critical point means that the number of visitors was ( ) in week ( ) at approximately ( ) visitors.\n(round to the nearest integer as needed.)
Answer
Explanation:
Step1: Analyze the derivative of (P(t))
First, find the derivative of (P(t)=121+(t - 88)e^{0.02t}) using the product rule ((uv)^\prime=u^\prime v+uv^\prime), where (u=t - 88) and (v = e^{0.02t}). (u^\prime=1), (v^\prime=0.02e^{0.02t}) (P^\prime(t)=e^{0.02t}+0.02(t - 88)e^{0.02t}=e^{0.02t}(1 + 0.02(t - 88))=e^{0.02t}(0.02t-1.76 + 1)=e^{0.02t}(0.02t - 0.76))
Step2: Find the critical point
Set (P^\prime(t)=0). Since (e^{0.02t}>0) for all (t), we solve (0.02t-0.76 = 0) (0.02t=0.76), so (t=\frac{0.76}{0.02}=38)
Step3: Determine the intervals of increase and decrease
- For (t\in(1,38)), let's take a test - value, say (t = 20). Then (P^\prime(20)=e^{0.02\times20}(0.02\times20 - 0.76)=e^{0.4}(0.4 - 0.76)<0). So (P(t)) is decreasing on ((1,38))
- For (t\in(38,52)), let's take a test - value, say (t = 40). Then (P^\prime(40)=e^{0.02\times40}(0.02\times40 - 0.76)=e^{0.8}(0.8 - 0.76)>0). So (P(t)) is increasing on ((38,52))
Step4: Interpret the critical point
When (t = 38), (P(38)=121+(38 - 88)e^{0.02\times38}=121-50e^{0.76}) (e^{0.76}\approx2.138), so (P(38)=121-50\times2.138=121 - 106.9\approx14)
Answer:
The critical point means that the number of visitors was (\text{minimum}) in week (38) at approximately (14) visitors.