Question 1155423
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)=x^2*sqrt(x+14)}}}.....derivate

apply the Product Rule: {{{(f* g )}}}'={{{f}}} '* {{{g}}}+{{{f}}}*{{{ g}}}'

{{{f¢(x)=(d/dx)x^2*sqrt(x+14)+x^2*(d/dx)(sqrt(x+14))}}}

{{{f¢(x)=2x*sqrt(x+14)+x^2*1/(2sqrt(x+14))}}}

{{{f¢(x)=(2x*sqrt(x+14)*2sqrt(x+14)+x^2)/(2sqrt(x+14))}}}

{{{f¢(x)=(4x*(x+14)+x^2)/(2sqrt(x+14))}}}

{{{f¢(x)=(4x^2+56+x^2)/(2sqrt(x+14))}}}

{{{f¢(x)=(5x^2+56)/(2sqrt(x+14))}}}

equate to zero and solve for {{{x}}}

{{{(5x^2+56)/(2sqrt(x+14))=0}}}-> will be zero if

{{{5x^2+56=0}}}
{{{5x^2=-56}}}
{{{x^2=-56/5}}}
{{{x=sqrt(-56/5)}}}-> complex solution, derivative is  undefined

if {{{(2sqrt(x+14))}}} derivative is  undefined

Critical points are points where the function is defined and its derivative is zero or undefined


{{{x= -56/5}}}, {{{x=0}}}, {{{x= -14}}}-> a critical points

Identify critical points not in f (x )  domain : {{{x>= -14}}

 domain of {{{f(x)}}} is  [{{{ -14}},{{{infinity}}})

{{{f(x)=x^2*sqrt(x+14)}}}
{{{f(x)=(-56/5)^2*sqrt(-56/5+14)}}}
{{{f(x)=209.9}}}
maximum is at: ({{{-14}}},{{{0}}})
{{{f(x)=x^2*sqrt(x+14)}}}
{{{f(x)=(-14)^2*sqrt(-14+14)}}}
{{{f(x)=(-14)^2*sqrt(0)}}}
{{{f(x)=0}}}
minimum is at: ({{{-14}}},{{{0}}})

{{{f(x)=0^2*sqrt(0+14)}}}
{{{f(x)=0}}}

minimum is at:  ({{{-14}}},{{{0}}}) and ({{{0}}},{{{0}}})

maximum  is at ({{{-56/5}}},{{{209.9}}}) 


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

({{{-14}}},{{{-56/5}}}) -> {{{f(x)}}} is increasing

({{{-56/5}}},{{{0}}})->{{{f(x)}}} is decreasing