Question 883627
Integer to find, n.  Is this for a Calculus course, or College Algebra/Intermediate Algebra?


{{{y=1/n+4n^2}}} and you want y as small as possible; and n must be an integer.  Use derivative, {{{dy/dn}}}.  This should be 0 at any extreme value for y, including any minimum.


{{{dy/dn=-n^(-2)+8n}}}
{{{8n-1/n^2}}}
{{{8n^3/n^2-1/n^2}}}
{{{(8n^3-1)/(n^2)}}}
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Finding extreme values for y
{{{(8n^3-1)=0}}}  ignoring the denominator because it has no importance for equating the derivative to zero.
Based on the difference of cubes formula, {{{highlight(((2n)-1)((2n)^2+2n+1)=0)}}}.


This would have three possible solutions for n, but maybe only one of them would be our minimum n for y.


The binomial:
{{{2n-1=0}}}
{{{2n=1}}}
{{{n=1/2}}}


The quadratic:
{{{4n^2+2n+1=0}}}
Discriminant, {{{2^2-4*4*1=4-16=-12<0}}}, not a real number.


Only that one, real, extreme value so must be our input for the minimum for y.
We want an INTEGER, so the nearest ones to {{{1/2}}} are <b>0 and 1</b>.
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In fact for smallest POSITIVE value for y, <b>choose n=1</b>.