Show that x2 + 4x + 17 is O(x3) but that x3 is not O(x2 +

Discrete Mathematics and Its Applications | 7th Edition | ISBN: 9780073383095 | Authors: Kenneth Rosen

ISBN: 9780073383095 37

Solution for problem 9E Chapter 3.2

Discrete Mathematics and Its Applications | 7th Edition

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Problem 9E

Show that $x^{2}+4 x+17$ is $O\left(x^{3}\right)$ but that $x^{3}$ is not $O\left(x^{2}+4 x+17\right)$

Step-by-Step Solution:

Step 1 of 2

Let us assume that x > 1 then ${x}^{3}{\ >x}^{2}\ >1$

So, the first term that is ${x}^{2}$ will influence.

Then, instead of ${x}^{2}$ we can write as ${x}^{3}$ . Because ${x}^{2}$ will grow as slow as ${x}^{3}$

And the second term is also an influence because x will grow as slow as ${x}^{2}.$

Hence,

${x}^{2}+4x+17\ =\ {x}^{3}\ +\ 4{x}^{3}\ +\ 17{x}^{3}\ =\ 22{x}^{3}$

Where n = 3, C = 22 and k = 1

So the growth of the function is O( ${x}^{3}$ ).

Hence, it is proved that ${x}^{2}+4x+17$ has the growth of function O( ${x}^{3}$ ).

Step 2 of 2

Chapter 3.2, Problem 9E is Solved

Textbook: Discrete Mathematics and Its Applications

Edition: 7

Author: Kenneth Rosen

ISBN: 9780073383095

Discrete Mathematics and Its Applications was written by and is associated to the ISBN: 9780073383095. This textbook survival guide was created for the textbook: Discrete Mathematics and Its Applications, edition: 7. The answer to “?Show that $x^{2}+4 x+17$ is $O\left(x^{3}\right)$ but that $x^{3}$ is not $O\left(x^{2}+4 x+17\right)$” is broken down into a number of easy to follow steps, and 13 words. This full solution covers the following key subjects: show. This expansive textbook survival guide covers 101 chapters, and 4221 solutions. Since the solution to 9E from 3.2 chapter was answered, more than 2006 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 9E from chapter: 3.2 was answered by , our top Math solution expert on 06/21/17, 07:45AM.