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## DIRECTIONS AND ANALYSIS

UNIT:FUNCTIONSDIRECTIONS AND ANALYSISTask 1: Saving for Vacation

Jerry plans to begin saving money for a trip by putting \$1 in a savings account the first month and then adding enough to double the amount in the account each following month.

After he finishes contributing to the account, he will withdraw \$500 to make a down payment on the trip. Create a function to show how much money will be in the account if he finishes contributing to the account after t months.

A tabulation of some the saving pattern.

Month Amount Deposited Total Balance

1 1 1

2 1 2

3 2 4

4 4 8

5 8 16

6 16 32

7 32 64

The total balance follows the pattern 1,2,4,8,16,32,64…

It behaves similar to a geometric sequence with common ratio r=2 and first term a=1.

An=arn-1At=1×2t-1 for t=1,2,3,4,5…At=2t-1 for t≥1Jerry plans to keep up this saving pattern until he can no longer afford it. At that time, he hopes to have enough in his account for his trip. How much will be in the account, after making the down payment, if the most he will be able to contribute in a month is \$1,200?

The maximum he can be able to contribute in a month is 1200 and at such a time the account will have 2×1200 since his deposit must double the amount.

Thus

At=2t-1=2×12002t-1=2400log2t-1=log2400t-1log2=log2400t-1=log2400log2t=1+log2400log2t=1+log2400log2t=1+11.23t=12.23Since at t=12.23 months he can add the maximum possible amount which is 1200 we need to round down to t=12 months to make sure he doesn’t surpass his ability.

Thus t= 12 months

A12=212-1=211=2048Balance after down payment

2048-500=1548Jerry’s friend Brandon is planning to start saving for a trip at the same time as Jerry. The graph shows the balance in Brandon’s account over the first few months of saving. Whose account will have the greatest balance after the first five months of savings, Brandon’s or Jerry’s?

Brandon saving pattern takes the form of a straight line

y=mx+cwhere the gradient m=∆y∆x=300-04-0=75and c=0ABt=75tAt t=5 months

AB5=75×5=375AJ5=25-1=24=16Thus Jerry’s account will have the greatest balance after the first five months of saving.

How long will it take for the person with the smaller account balance after five months to have the greater balance?

We need to get the time that Jerry will require to achieve Brandon’s amount at 5 months.

AJt=2t-1=375log2t-1=log375t-1log2=log375(t-1)=log375log2t=1+log375log2t=1+8.55t=9.55 monthsBrandon will also need to withdraw \$500 to make a down payment on the trip. How much will be in Brandon’s account, after making the down payment, if he contributes for same amount of time as Jerry? (Review your calculation in part b to determine how long Jerry continued his savings plan.)

Amount for Brandon after 12 months.

ABt=75tAB12=75×12AB12=900Balance after down payment

Jerry is driving to his vacation destination. He decided to keep track of the distance to his destination after different amounts of times spent driving. He created a table from the data he collected.

Hours Spent Driving 1 5 6 9

Miles to Destination 1,259 999 934 739

Assuming that the pattern is the same for Jerry’s entire drive to his destination, create a function to represent the distance remaining to his destination based on the number of hours, h, he has driven.

f1=1259;f5=999;f6=934;f9=739Rate of change of distance per hour for the different intervals.

for 1-5 hrs=999-12595-1=-65for 5-6 hrs=934-9996-5=-65for 6-9 hrs=739-9349-6=-65Clearly the modelling function will be a straight line of gradient m=-65

y=mx+cWhere c is the total distance from origin to destination which occurs at h=0.

From 5-6 hrs it can be observed he covers 999-934=65 miles in an hour.

At 1 hr distance =1259

Thus

total distance=1259+65=1324 miles∴fh=-65h+1324What is the domain of this function in the context of the problem? Explain.

The domain will be the set of values which when plugged into the function will give the reasonable outputs between 0 to 1324 miles which the distance from Jerry’s origin to destination.

And

h≥0 since there are no negative hoursThe upper bound for the domain will be the time he takes to cover the whole journey when he gets at the destination.

fh=0=-65h+1324-65h=-1324h=-1324-65h=132465∴The domain is:0≤h≤132465What is the range of this function in the context of the problem? Explain.

This the set all possible outputs of the function for the given domain.

f0=-650+1324=1324f132465=-65132465+1324=-1324+1324=0∴The range is:0≤f(h)≤1324This the range of the distance from the origin to the destination.

When Jerry recorded the last entry in the table (9 hours driven, 739 miles to go), how much longer did he need to drive to arrive at his destination?

At the destination

fh=0=-65h+1324-65h=-1324h=-1324-65=132465 hrsHe needed

132465-9=73965hrs to complete the journey≈11.369 hrs(3 decimal places)Resources HYPERLINK “https://www.varsitytutors.com/hotmath/hotmath_help/topics/domain-and-range”

Varsity Tutors, “Domain and Range,” https://www.varsitytutors.com/hotmath/hotmath_help/topics/domain-and-range.

James Stewart, Essential Calculus, 2nd edition, Cengage Learning, 2012.

UNIT:FUNCTIONSDIRECTIONS AND ANALYSISTask 1: Saving for Vacation

Jerry plans to begin saving money for a trip by putting \$1 in a savings account the first month and then adding enough to double the amount in the account each following month.

After he finishes contributing to the account, he will withdraw \$500 to make a down payment on the trip. Create a function to show how much money will be in the account if he finishes contributing to the account after t months.

A tabulation of some the saving pattern.

Month Amount Deposited Total Balance

1 1 1

2 1 2

3 2 4

4 4 8

5 8 16

6 16 32

7 32 64

The total balance follows the pattern 1,2,4,8,16,32,64…

It behaves similar to a geometric sequence with common ratio r=2 and first term a=1.

An=arn-1At=1×2t-1 for t=1,2,3,4,5…At=2t-1 for t≥1Jerry plans to keep up this saving pattern until he can no longer afford it. At that time, he hopes to have enough in his account for his trip. How much will be in the account, after making the down payment, if the most he will be able to contribute in a month is \$1,200?

The maximum he can be able to contribute in a month is 1200 and at such a time the account will have 2×1200 since his deposit must double the amount.

Thus

At=2t-1=2×12002t-1=2400log2t-1=log2400t-1log2=log2400t-1=log2400log2t=1+log2400log2t=1+log2400log2t=1+11.23t=12.23Since at t=12.23 months he can add the maximum possible amount which is 1200 we need to round down to t=12 months to make sure he doesn’t surpass his ability.

Thus t= 12 months

A12=212-1=211=2048Balance after down payment

2048-500=1548Jerry’s friend Brandon is planning to start saving for a trip at the same time as Jerry. The graph shows the balance in Brandon’s account over the first few months of saving. Whose account will have the greatest balance after the first five months of savings, Brandon’s or Jerry’s?

Brandon saving pattern takes the form of a straight line

y=mx+cwhere the gradient m=∆y∆x=300-04-0=75and c=0ABt=75tAt t=5 months

AB5=75×5=375AJ5=25-1=24=16Thus Jerry’s account will have the greatest balance after the first five months of saving.

How long will it take for the person with the smaller account balance after five months to have the greater balance?

We need to get the time that Jerry will require to achieve Brandon’s amount at 5 months.

AJt=2t-1=375log2t-1=log375t-1log2=log375(t-1)=log375log2t=1+log375log2t=1+8.55t=9.55 monthsBrandon will also need to withdraw \$500 to make a down payment on the trip. How much will be in Brandon’s account, after making the down payment, if he contributes for same amount of time as Jerry? (Review your calculation in part b to determine how long Jerry continued his savings plan.)

Amount for Brandon after 12 months.

ABt=75tAB12=75×12AB12=900Balance after down payment

Jerry is driving to his vacation destination. He decided to keep track of the distance to his destination after different amounts of times spent driving. He created a table from the data he collected.

Hours Spent Driving 1 5 6 9

Miles to Destination 1,259 999 934 739

Assuming that the pattern is the same for Jerry’s entire drive to his destination, create a function to represent the distance remaining to his destination based on the number of hours, h, he has driven.

f1=1259;f5=999;f6=934;f9=739Rate of change of distance per hour for the different intervals.

for 1-5 hrs=999-12595-1=-65for 5-6 hrs=934-9996-5=-65for 6-9 hrs=739-9349-6=-65Clearly the modelling function will be a straight line of gradient m=-65

y=mx+cWhere c is the total distance from origin to destination which occurs at h=0.

From 5-6 hrs it can be observed he covers 999-934=65 miles in an hour.

At 1 hr distance =1259

Thus

total distance=1259+65=1324 miles∴fh=-65h+1324What is the domain of this function in the context of the problem? Explain.

The domain will be the set of values which when plugged into the function will give the reasonable outputs between 0 to 1324 miles which the distance from Jerry’s origin to destination.

And

h≥0 since there are no negative hoursThe upper bound for the domain will be the time he takes to cover the whole journey when he gets at the destination.

fh=0=-65h+1324-65h=-1324h=-1324-65h=132465∴The domain is:0≤h≤132465What is the range of this function in the context of the problem? Explain.

This the set all possible outputs of the function for the given domain.

f0=-650+1324=1324f132465=-65132465+1324=-1324+1324=0∴The range is:0≤f(h)≤1324This the range of the distance from the origin to the destination.

When Jerry recorded the last entry in the table (9 hours driven, 739 miles to go), how much longer did he need to drive to arrive at his destination?

At the destination

fh=0=-65h+1324-65h=-1324h=-1324-65=132465 hrsHe needed

132465-9=73965hrs to complete the journey≈11.369 hrs(3 decimal places)Resources HYPERLINK “https://www.varsitytutors.com/hotmath/hotmath_help/topics/domain-and-range”

Varsity Tutors, “Domain and Range,” https://www.varsitytutors.com/hotmath/hotmath_help/topics/domain-and-range.

James Stewart, Essential Calculus, 2nd edition, Cengage Learning, 2012.

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