PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)
PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)
8th Edition
ISBN: 9781337904285
Author: Larson
Publisher: CENGAGE L
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Chapter 3.2, Problem 111E

a.

To determine

To complete:

The given table and interpret your results.

a.

Expert Solution
Check Mark

Answer to Problem 111E

The complete table would be:

    K 1 2 4 6 8 10 12
    t 0 19.80 39.6151.1959.4165.79 71.00

Explanation of Solution

Given:

A principalP , invested at 312% and compounded continuously, increases to an amount K times the original principal after t years, where T=ln(K)0.035.

    K 1 2 4 6 8 10 12
    t

Calculation:

To complete the given table, we will substitute the values of K in our given function as shown below:

  T=ln(K)0.035=ln(1)0.035=00.035=0

  T=ln(K)0.035=ln(2)0.035=0.69314718055990.035=19.8042051588≈19.80

It will take approximately 19.80 years for the investment to double.

  T=ln(K)0.035=ln(4)0.035=1.386294361120.035=39.6084103177≈39.61

It will take approximately 39.61 years for the investment to be 4 times original amount.

  T=ln(K)0.035=ln(6)0.035=1.791759469230.035=51.1931276922≈51.19

It will take approximately 51.19 years for the investment to be 6 times original amount.

  T=ln(K)0.035=ln(8)0.035=2.079441541680.035=59.4126154766≈59.41

It will take approximately 59.41 years for the investment to be 8 times original amount.

  T=ln(K)0.035=ln(10)0.035=2.302585092990.035=65.7881455141≈65.79

It will take approximately 65.79 years for the investment to be 10 times original amount.

  T=ln(K)0.035=ln(12)0.035=2.484906649790.035=70.9973328511≈71.00

It will take approximately 71.00 years for the investment to be 12 times original amount.

Therefore, the complete table would be:

    K 1 2 4 6 8 10 12
    t 0 19.80 39.61 51.19 59.41 65.79 71.00

b.

To determine

To graph:

The given function using graphing utility.

b.

Expert Solution
Check Mark

Answer to Problem 111E

The graph of our given function would be:

  PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS), Chapter 3.2, Problem 111E , additional homework tip 1

Explanation of Solution

Given:

A principal P , invested at 312% and compounded continuously, increases to an amount K times the original principal after t years, where T=ln(K)0.035.

Calculation:

Upon graphing our given function, we will get our required graph as shown below:

  PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS), Chapter 3.2, Problem 111E , additional homework tip 2

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Chapter 3 Solutions

PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)

Ch. 3.1 - Prob. 1ECh. 3.1 - Prob. 2ECh. 3.1 - Prob. 3ECh. 3.1 - Prob. 4ECh. 3.1 - Prob. 5ECh. 3.1 - Prob. 6ECh. 3.1 - Prob. 7ECh. 3.1 - Prob. 8ECh. 3.1 - Prob. 9ECh. 3.1 - Prob. 10E
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Prob. 32ECh. 3.6 - Prob. 33ECh. 3.6 - Prob. 34ECh. 3.6 - Prob. 35ECh. 3.6 - Prob. 36ECh. 3.6 - Prob. 37ECh. 3.6 - Prob. 38ECh. 3.6 - Prob. 39ECh. 3.6 - Prob. 40ECh. 3.6 - Prob. 41ECh. 3.6 - Prob. 42ECh. 3.6 - Prob. 43ECh. 3.6 - Prob. 44ECh. 3.6 - Prob. 45ECh. 3.6 - Prob. 46ECh. 3 - Prob. 1CRCh. 3 - Prob. 2CRCh. 3 - Prob. 3CRCh. 3 - Prob. 4CRCh. 3 - Prob. 5CRCh. 3 - Prob. 6CRCh. 3 - Prob. 7CRCh. 3 - Prob. 8CRCh. 3 - Prob. 9CRCh. 3 - Prob. 10CRCh. 3 - Prob. 11CRCh. 3 - Prob. 12CRCh. 3 - Prob. 13CRCh. 3 - Prob. 14CRCh. 3 - Prob. 15CRCh. 3 - Prob. 16CRCh. 3 - Prob. 17CRCh. 3 - Prob. 18CRCh. 3 - Prob. 19CRCh. 3 - Prob. 20CRCh. 3 - Prob. 21CRCh. 3 - Prob. 22CRCh. 3 - Prob. 23CRCh. 3 - Prob. 24CRCh. 3 - Prob. 25CRCh. 3 - Prob. 26CRCh. 3 - Prob. 27CRCh. 3 - Prob. 28CRCh. 3 - Prob. 29CRCh. 3 - Prob. 30CRCh. 3 - Prob. 31CRCh. 3 - Prob. 32CRCh. 3 - Prob. 33CRCh. 3 - Prob. 34CRCh. 3 - Prob. 35CRCh. 3 - Prob. 36CRCh. 3 - Prob. 37CRCh. 3 - Prob. 38CRCh. 3 - Prob. 39CRCh. 3 - Prob. 40CRCh. 3 - Prob. 41CRCh. 3 - Prob. 42CRCh. 3 - Prob. 43CRCh. 3 - Prob. 44CRCh. 3 - Prob. 45CRCh. 3 - Prob. 46CRCh. 3 - Prob. 47CRCh. 3 - Prob. 48CRCh. 3 - Prob. 49CRCh. 3 - Prob. 50CRCh. 3 - Prob. 51CRCh. 3 - Prob. 52CRCh. 3 - Prob. 53CRCh. 3 - Prob. 54CRCh. 3 - Prob. 55CRCh. 3 - Prob. 56CRCh. 3 - Prob. 57CRCh. 3 - Prob. 58CRCh. 3 - Prob. 59CRCh. 3 - Prob. 60CRCh. 3 - Prob. 61CRCh. 3 - Prob. 62CRCh. 3 - Prob. 63CRCh. 3 - Prob. 64CRCh. 3 - Prob. 65CRCh. 3 - Prob. 66CRCh. 3 - Prob. 67CRCh. 3 - Prob. 68CRCh. 3 - Prob. 69CRCh. 3 - Prob. 70CRCh. 3 - Prob. 71CRCh. 3 - Prob. 72CRCh. 3 - Prob. 73CRCh. 3 - Prob. 74CRCh. 3 - Prob. 75CRCh. 3 - Prob. 76CRCh. 3 - Prob. 77CRCh. 3 - Prob. 78CRCh. 3 - Prob. 79CRCh. 3 - Prob. 80CRCh. 3 - Prob. 81CRCh. 3 - Prob. 82CRCh. 3 - Prob. 83CRCh. 3 - Prob. 84CRCh. 3 - Prob. 85CRCh. 3 - Prob. 86CRCh. 3 - Prob. 87CRCh. 3 - Prob. 88CRCh. 3 - Prob. 89CRCh. 3 - Prob. 90CRCh. 3 - Prob. 91CRCh. 3 - Prob. 92CRCh. 3 - Prob. 93CRCh. 3 - Prob. 94CRCh. 3 - Prob. 95CRCh. 3 - Prob. 96CRCh. 3 - Prob. 97CRCh. 3 - Prob. 98CRCh. 3 - Prob. 99CRCh. 3 - Prob. 100CRCh. 3 - Prob. 101CRCh. 3 - Prob. 102CRCh. 3 - Prob. 103CRCh. 3 - Prob. 104CRCh. 3 - Prob. 105CRCh. 3 - Prob. 106CRCh. 3 - Prob. 107CRCh. 3 - Prob. 108CRCh. 3 - Prob. 109CRCh. 3 - Prob. 110CRCh. 3 - Prob. 111CRCh. 3 - Prob. 112CRCh. 3 - Prob. 113CRCh. 3 - Prob. 114CRCh. 3 - Prob. 115CRCh. 3 - Prob. 116CRCh. 3 - Prob. 117CRCh. 3 - Prob. 118CRCh. 3 - Prob. 119CRCh. 3 - Prob. 120CRCh. 3 - Prob. 121CRCh. 3 - Prob. 122CRCh. 3 - Prob. 123CRCh. 3 - Prob. 124CRCh. 3 - Prob. 125CRCh. 3 - Prob. 126CRCh. 3 - Prob. 1CTCh. 3 - Prob. 2CTCh. 3 - Prob. 3CTCh. 3 - Prob. 4CTCh. 3 - Prob. 5CTCh. 3 - Prob. 6CTCh. 3 - Prob. 7CTCh. 3 - Prob. 8CTCh. 3 - Prob. 9CTCh. 3 - Prob. 10CTCh. 3 - Prob. 11CTCh. 3 - Prob. 12CTCh. 3 - Prob. 13CTCh. 3 - Prob. 14CTCh. 3 - Prob. 15CTCh. 3 - Prob. 16CTCh. 3 - Prob. 17CTCh. 3 - Prob. 18CTCh. 3 - Prob. 19CTCh. 3 - Prob. 20CTCh. 3 - Prob. 21CTCh. 3 - Prob. 22CTCh. 3 - Prob. 23CTCh. 3 - Prob. 24CTCh. 3 - Prob. 25CTCh. 3 - Prob. 26CTCh. 3 - Prob. 27CTCh. 3 - Prob. 28CTCh. 3 - Prob. 29CTCh. 3 - Prob. 30CTCh. 3 - Prob. 1STPCh. 3 - Prob. 2STPCh. 3 - Prob. 3STPCh. 3 - Prob. 4STPCh. 3 - Prob. 5STPCh. 3 - Prob. 6STPCh. 3 - Prob. 7STPCh. 3 - Prob. 8STPCh. 3 - Prob. 9STPCh. 3 - Prob. 10STPCh. 3 - Prob. 11STPCh. 3 - Prob. 12STPCh. 3 - Prob. 13STPCh. 3 - Prob. 14STPCh. 3 - Prob. 15STPCh. 3 - Prob. 16STPCh. 3 - Prob. 17STPCh. 3 - Prob. 18STPCh. 3 - Prob. 19STPCh. 3 - Prob. 20STPCh. 3 - Prob. 21STPCh. 3 - Prob. 22STPCh. 3 - Prob. 23STPCh. 3 - Prob. 24STPCh. 3 - Prob. 25STPCh. 3 - Prob. 26STP
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Implicit Differentiation with Transcendental Functions; Author: Mathispower4u;https://www.youtube.com/watch?v=16WoO59R88w;License: Standard YouTube License, CC-BY
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