Question 1192429

Extreme Points of {{{y = x^2/(x+1)}}}:

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}}}. 


find {{{f}}} '{{{(x )}}}


{{{y = x^2/(x+1)}}}..........{{{y=f(x)}}}

{{{f(x)= x^2/(x+1)}}}

Apply the Quotient Rule: {{{(f/g)}}}'=({{{f}}}'{{{(g)}}}-{{{f}}}({{{g}}}'))/{{{g^2}}}


{{{f}}}'{{{(x) =(2x(x+1)-x^2*1)/(x+1)^2}}}

{{{f}}}'{{{(x) =(2x^2+2x-x^2)/(x+1)^2}}}

{{{f}}}'{{{(x) =(x^2+2x)/(x+1)^2}}}

{{{f}}}'{{{(x) = (x (x + 2))/(x + 1)^2}}}


{{{f}}}'{{{(x) =0}}}


{{{(x (x + 2))/(x + 1)^2=0}}} will be only  if numerator equal to zero, so

{{{x (x + 2)=0}}}

=>{{{x=0 }}}or {{{x=-2}}}


then

{{{f(0)= 0^2/(0+1)=0}}}-> point is ({{{0}}},{{{0}}})

{{{f(-2)= (-2)^2/(-2+1)=4/-1=-4}}}-> point is ({{{-2}}},{{{-4}}})



so we have:

Maximum:  at ({{{-2}}},{{{-4}}})
Minimum: at ({{{0}}},{{{0}}})


{{{ drawing( 600, 600, -10, 10, -10, 10,
circle(-2,-4,.12), locate(-2,-4,max(-2,-4)),locate(0.3,-0.3,min(0,0)),
graph( 600, 600, -10, 10, -10, 10,x^2/(x+1))) }}}