Question 1155422

Suppose that x=c is a critical point of {{{f (x )}}}.
If {{{f}}}'{{{(x )>0 }}}to the left of  {{{x=c}}}  and  {{{f}}}'{{{(x )<0}}} to the right of  {{{x=c}}}, then  {{{x=c}}} is a local {{{maximum}}}.
If {{{f}}}'{{{(x )<0 }}}to the left of  {{{x=c}}}  and  {{{f}}}'{{{(x )>0}}} to the right of  {{{x=c}}}, then  {{{x=c}}} is a local {{{minimum}}}.
If {{{f}}}'{{{(x )}}}  is the {{{same}}}{{{ sign}}} on both sides of {{{x=c}}} then {{{x=c}}} is neither a local maximum nor a local minimum. 


{{{f(x)=2x^2-3x-5}}}.....derivate

{{{f}}}'{{{(x)=4x-3}}}

{{{4x-3=0}}}

{{{x=3/4}}}-> a critical point: minimum

 domain of {{{f(x)}}} is 

({{{-infinity}}},{{{infinity}}})

{{{f(x)}}} is monotone in intervals:

{{{-infinity<x<3/4}}} -> {{{f(x)}}} is decreasing

{{{3/4<x<infinity}}}->{{{f(x)}}} is increasing



{{{f(x)=2(3/4)^2-3(3/4)-5}}}
{{{f(x)=2(9/16)-3(3/4)-5}}}
{{{f(x)=(9/8)-(9/4)-5}}}
{{{f(x)=9/8-18/8-40/8}}}
{{{f(x)=-9/8-40/8}}}
{{{f(x)=-49/8}}}

minimum is at: ({{{3/4}}},{{{-49/8}}})