Find the least integer n such that f(x) is O(xn) for each of these functions. a) f(x)=2x2+x3log⁡x b) f(x)=3x5+(log⁡x)4 c) f(x)=x4+x2+1x4+1

Question

Find the least integer n such that f(x) is O(xn) for each of these functions.  a) f(x)=2x2+x3log⁡x  b) f(x)=3x5+(log⁡x)4  c) f(x)=x4+x2+1x4+1

Toni Scott · Accepted Answer

a) Given:  f(x)=2x2+x3log⁡x  When x&amp;gt;0, we have the property log⁡x≤x  f(x)=2x2+x3log⁡x≤2x2+x4  The largest power of x mentioned in the definition of f(x) is the smallest n for which f(x) is O(xn)  n=4  When x&amp;gt;2, we have the property x2&amp;gt;x&amp;gt;2  For convience sake, we will choose k=2 and thus use x&amp;gt;2  |f(x)|=|2x2+x3log⁡x|≤|2x2+x4|  ≤|2x2|+|x4|  =2x2+x4≤x4+x4  =2x4  =2|x4|  Thus we need to choose C to be at least 2. Let us then take C=2  By the definition of the Big-O notation, f(x)=O(x4) with k=2 and C=2

maul124uk · Accepted Answer

b) Given:  f(x)=3x5+(log⁡x)4  The largest power of x mentioned in the definition of f(x) is the smallest n for which f(x) is O(xn)  n=5  When x.0, we have the property log⁡x≤x  When x&amp;gt;1, we also have the property x4&amp;lt;x5  For convience sake, we will choose k=1 and thus us x&amp;gt;1  |f(x)|=|3x5+(log⁡x)4|≤|3x5|+|(log⁡x)4|  =3x5+(log⁡x)4≤3x5+x4  =4x5  =4|x5|  Thus we nned to choose C to be at least 4. Let us then take C=4  By the definition of the Big-O notation f(x)=O(x5) with k=1 and C=4

RizerMix · Accepted Answer

c) Given: f(x)=x4+x2+1x4+1 Let us determine the quotient (with remainder) of the given using long division. The quotient is then 1 with remainder x2 and thus we can rewrite the given fractions as: f(x)=x4+x2+1x4+1=1+x2x4+1 The largest power of x of the quotient is the smallest n for which f(x) is O(xn). n=0 For convenience sake, we will choose k=0 and thus x&amp;gt;0 |f(x)=|1+x2x4+1|≤|1|+|x2x4+1| =1+x2x4+1≤1+1 &amp;lt;2 =2⋅|1| =2|x0| Thus we need to choose C to be at least 2. Let us then take C=2 By the definition of the Big-O notation. f(x)=O(x0)=O(1) with k=0 and C=2

karton · Accepted Answer

Answer:a) n=3b) n=5c) n=4Explanation:To find the least integer n such that f(x) is O(xn) for each of the given functions, we can analyze the highest degree terms in the functions.a) For f(x)=2x2+x3log(x), the highest degree term is x3log(x). Since log(x) grows slower than any positive power of x, we can ignore it for our analysis. Therefore, the highest degree term is x3, and we can express f(x) as f(x)=2x2+O(x3). Hence, n=3.b) For f(x)=3x5+(log(x))4, the highest degree term is x5. The logarithmic term, (log(x))4, grows slower than any positive power of x. Thus, we can express f(x) as f(x)=3x5+O(x5). Therefore, n=5.c) For f(x)=x4+x2+1x4+1, both the numerator and denominator have the highest degree of x4. When x tends to infinity, the other terms become negligible compared to the leading term. Thus, we can express f(x) as f(x)=x4x4+O(x4)=1+O(x4). Therefore, n=4.

star233 · Accepted Answer

a) For the function f(x)=2x2+x3logx, we need to find the least integer n such that f(x) is O(xn).Let&#039;s analyze the growth of each term separately:f(x)=2x2+x3logx=O(x2)+O(x3logx)Since x3logx grows faster than x2, we can ignore the O(x2) term. Therefore, we have:f(x)=O(x3logx)So, the least integer n such that f(x) is O(xn) is n=3.b) For the function f(x)=3x5+(logx)4, we need to find the least integer n such that f(x) is O(xn).Similar to the previous example, let&#039;s analyze the growth of each term:f(x)=3x5+(logx)4=O(x5)+O((logx)4)Since (logx)4 grows slower than x5, we can ignore the O((logx)4) term. Therefore, we have:f(x)=O(x5)Hence, the least integer n such that f(x) is O(xn) is n=5.c) For the function f(x)=x4+x2+1x4+1, we need to find the least integer n such that f(x) is O(xn).Let&#039;s simplify the expression:f(x)=x4+x2+1x4+1=1+1x2+1x41+1x4As x approaches infinity, the terms 1x2 and 1x4 become negligible compared to 1. Thus, we have:f(x)=1+0+01+0=1Since f(x) is a constant value, it is O(x0) for any positive value of x. Therefore, the least integer n such that f(x) is O(xn) is n=0.To summarize:a) n=3b) n=5c) n=0

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