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Given the following four functions $f_{1}(n)=n^{100},$  $f_{2}(n)=(1.2)^{n},$  $f_{3}(n)=2^{n/2},$ $f_{4}(n)=3^{n/3}$ which function will have the largest value for sufficiently large values of n $(i.e.$ $n\rightarrow\infty)?$

  1. $f_{4}$
  2. $f_{3}$
  3. $f_{2}$
  4. $f_{1}$
in Numerical Ability by Boss (47.5k points)
edited by | 36 views

1 Answer

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Best answer
  • $f_{1}(n) = n^{100}$
  • $f_{2}(n) = (1.2)^{n}$
  • $f_{3}(n) = (2)^{n/2} = (1.414)^{n}$
  • $f_{4}(n) = (3)^{n/3} = (1.442)^{n}$

We know that $(1.442)^{n} > (1.414)^{n}$ and $(1.442)^{n} > (1.2)^{n}$

$\Rightarrow$$f_{4}(n)$ is greater than both $f_{2}(n)$ and $f_{3}(n)$

Now we have to check between $f_{4}(n)$ and $f_{1}(n)$

But $f_1(n)$ is a polynomial function and $f_4(n)$ is a exponential function. Any exponential function will have larger value than any polynomial function for sufficiently large value of input. So, $f_4(n)$ has the largest value here.

Option A. $f_{4}(n)$ is the correct answer.

by Boss (18.3k points)
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