Question 979130
1. Find the derivative.
{{{df/dx=2x-6}}}
Set it equal to zero.
{{{2x-6=0}}}
{{{2x=6}}}
{{{x=3}}}
So when {{{x=3}}},
{{{y=3^2-6(3)+2}}}
{{{y=9-18+2}}}
{{{y=7}}}
Since the second derivative, {{{2>0}}}, the extrema at {{{x=3}}} is a minimum.
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2. {{{lim(x->4,(x^2-16)/(x-4))=lim(x->4,(x+4))=4+4=8}}} - Removable discontinuity.
{{{lim(x->4,(x^2-16)/(x-4))=lim(x->4,(2x)/(1))=2(4)=8}}} - L'Hopital's Rule 
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3. Function in red, second derivative is in purple.
Function (red) is concave up when the second derivative (purple) is greater than zero.
*[illustration a56.JPG].