Modeling the Dynamics of Life: Calculus and Probability for Life Scientists
Modeling the Dynamics of Life: Calculus and Probability for Life Scientists
3rd Edition
ISBN: 9780840064189
Author: Frederick R. Adler
Publisher: Cengage Learning
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Chapter 2.3, Problem 27E

(a)

To determine

To sketch:The graph of the continuous function.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given information:

A continuous function that is 1forx0.1,1for x0.1 and is linear for 0.1<x<0.1.

Graph:

The sigma function is defined as the function which has the value -1for the negative argument and 0 when the argument is 0 and 1 for the positive argument.

The graph for the given continuous function 1forx0.1,1for x0.1 and is linear for 0.1<x<0.1. is as follows,

  Modeling the Dynamics of Life: Calculus and Probability for Life Scientists, Chapter 2.3, Problem 27E

Interpretation:

From the graph it is obtained that the value -1 has negative argument and 1 has positive argument. The graph is increasing and decreasing with respect to the argument value.

(b)

To determine

To find:The formula for the function.

(b)

Expert Solution
Check Mark

Answer to Problem 27E

The formula for the function is given below as,

  S(x)={1ifx0.11+10(x+0.1)if-0.1<x<0.11if x0.1}

Explanation of Solution

Given information:

A continuous function that is 1forx0.1,1for x0.1 and is linear for 0.1<x<0.1.

Calculation:

To determine the formula for the function as follows,

Consider the slope for (0.1.l) as 10, and the linear function for 0.1<x<0.1 .Linear function is written in the form of f(x)=mx+b .Here the value of m=10 and as we want the linear for 0.1<x<0.1 .Hence the function as 1+10(x+0.1) .

The formula for the function is given below as,

  S(x)={1ifx0.11+10(x+0.1)if-0.1<x<0.11if x0.1}

(c)

To determine

The input have to be to 0 for the output to be within 0.1 for 0.

(c)

Expert Solution
Check Mark

Answer to Problem 27E

The input should be within -0.01 to 0.01 for the output to be within 0.1 to 0.

Explanation of Solution

Given information:

A continuous function that is 1forx0.1,1for x0.1 and is linear for 0.1<x<0.1.

Calculation:

The given function is f(x)=1+10(x+0.1) .Since it is polynomial, it is continuous.

To determine the input at x=0 for the output within 0.1 to 0, first suppose that f(x)=0.1 implies x=0.01 .

Similarly, consider f(x)=0.1 implies x=0.01

From the two values of f(x) we have,

  0.11+10(x+0.1)0.10.110x0.10.01x0.01

Hence, the input should be within -0.01 to 0.01 for the output to be within 0.1 to 0.

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Chapter 2 Solutions

Modeling the Dynamics of Life: Calculus and Probability for Life Scientists

Ch. 2.1 - Prob. 11ECh. 2.1 - Prob. 12ECh. 2.1 - Prob. 13ECh. 2.1 - Prob. 14ECh. 2.1 - Prob. 15ECh. 2.1 - Prob. 16ECh. 2.1 - Prob. 17ECh. 2.1 - Prob. 18ECh. 2.1 - Prob. 19ECh. 2.1 - Prob. 20ECh. 2.1 - Prob. 21ECh. 2.1 - Prob. 22ECh. 2.1 - Prob. 23ECh. 2.1 - Prob. 24ECh. 2.1 - Prob. 25ECh. 2.1 - Prob. 26ECh. 2.1 - Prob. 27ECh. 2.1 - Prob. 28ECh. 2.1 - Prob. 29ECh. 2.1 - Prob. 30ECh. 2.1 - Prob. 31ECh. 2.1 - Prob. 32ECh. 2.1 - Prob. 33ECh. 2.1 - Prob. 34ECh. 2.1 - Prob. 35ECh. 2.1 - Prob. 36ECh. 2.1 - Prob. 37ECh. 2.1 - Prob. 38ECh. 2.1 - Prob. 39ECh. 2.1 - Prob. 40ECh. 2.1 - Prob. 41ECh. 2.1 - Prob. 42ECh. 2.1 - Prob. 43ECh. 2.1 - Prob. 44ECh. 2.1 - Prob. 45ECh. 2.1 - Prob. 46ECh. 2.2 - Prob. 1ECh. 2.2 - Prob. 2ECh. 2.2 - Prob. 3ECh. 2.2 - Prob. 4ECh. 2.2 - Prob. 5ECh. 2.2 - Prob. 6ECh. 2.2 - Prob. 7ECh. 2.2 - Prob. 8ECh. 2.2 - Prob. 9ECh. 2.2 - Prob. 10ECh. 2.2 - Prob. 11ECh. 2.2 - Prob. 12ECh. 2.2 - Prob. 13ECh. 2.2 - Prob. 14ECh. 2.2 - Prob. 15ECh. 2.2 - Prob. 16ECh. 2.2 - Prob. 17ECh. 2.2 - 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Prob. 24ECh. 2.9 - Prob. 25ECh. 2.9 - Prob. 26ECh. 2.9 - Prob. 27ECh. 2.9 - Prob. 28ECh. 2.9 - Prob. 29ECh. 2.9 - Prob. 30ECh. 2.9 - Prob. 31ECh. 2.9 - Prob. 32ECh. 2.9 - Prob. 33ECh. 2.9 - Prob. 34ECh. 2.9 - Prob. 35ECh. 2.9 - Prob. 36ECh. 2.9 - Prob. 38ECh. 2.9 - Prob. 39ECh. 2.9 - Prob. 40ECh. 2.9 - Prob. 41ECh. 2.9 - Prob. 42ECh. 2.9 - Prob. 43ECh. 2.9 - Prob. 44ECh. 2.9 - The method of implicit differentiation is often...Ch. 2.9 - Prob. 46ECh. 2.9 - Prob. 47ECh. 2.9 - Prob. 48ECh. 2.9 - Prob. 49ECh. 2.9 - Prob. 50ECh. 2.9 - Prob. 51ECh. 2.9 - Prob. 52ECh. 2.9 - Prob. 53ECh. 2.10 - Prob. 1ECh. 2.10 - Prob. 2ECh. 2.10 - Prob. 3ECh. 2.10 - Prob. 4ECh. 2.10 - Prob. 5ECh. 2.10 - Prob. 6ECh. 2.10 - Prob. 7ECh. 2.10 - Prob. 8ECh. 2.10 - Prob. 9ECh. 2.10 - Prob. 10ECh. 2.10 - Prob. 11ECh. 2.10 - Prob. 12ECh. 2.10 - Prob. 13ECh. 2.10 - Prob. 14ECh. 2.10 - Prob. 15ECh. 2.10 - Prob. 16ECh. 2.10 - Prob. 17ECh. 2.10 - Prob. 18ECh. 2.10 - Prob. 19ECh. 2.10 - Prob. 20ECh. 2.10 - Prob. 21ECh. 2.10 - Prob. 22ECh. 2.10 - Prob. 23ECh. 2.10 - Prob. 24ECh. 2.10 - Prob. 25ECh. 2.10 - Prob. 26ECh. 2.10 - Prob. 27ECh. 2.10 - Prob. 28ECh. 2.10 - Prob. 29ECh. 2.10 - Prob. 30ECh. 2.10 - Prob. 31ECh. 2.10 - Prob. 32ECh. 2.10 - Prob. 33ECh. 2.10 - Prob. 34ECh. 2.10 - Prob. 35ECh. 2.10 - Prob. 36ECh. 2.10 - Prob. 37ECh. 2.10 - Prob. 38ECh. 2.10 - Prob. 39ECh. 2.10 - Prob. 40ECh. 2.10 - Prob. 41ECh. 2.10 - Prob. 42ECh. 2.10 - Prob. 43ECh. 2 - Prob. 1SPCh. 2 - Prob. 2SPCh. 2 - Prob. 3SPCh. 2 - Prob. 4SPCh. 2 - Prob. 5SPCh. 2 - Prob. 6SPCh. 2 - Prob. 7SPCh. 2 - Prob. 8SPCh. 2 - Prob. 9SPCh. 2 - Prob. 10SPCh. 2 - Prob. 11SPCh. 2 - Prob. 12SPCh. 2 - Prob. 13SPCh. 2 - Prob. 14SPCh. 2 - Prob. 15SPCh. 2 - Prob. 16SPCh. 2 - Prob. 17SPCh. 2 - Prob. 18SPCh. 2 - Prob. 19SPCh. 2 - Prob. 20SPCh. 2 - Prob. 21SPCh. 2 - Prob. 22SPCh. 2 - Prob. 23SPCh. 2 - Prob. 24SPCh. 2 - Prob. 25SPCh. 2 - Prob. 26SPCh. 2 - Prob. 27SPCh. 2 - Prob. 28SPCh. 2 - Prob. 29SPCh. 2 - Prob. 30SPCh. 2 - Prob. 31SPCh. 2 - Prob. 32SPCh. 2 - 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