How do use the first derivative test to determine the local extrema #(x)/[(x^2) +3]#?

1 Answer

Answer:

Refer to explanation

Explanation:

The First Derivative Test states

Suppose #f(x)# is continuous at a critical point #x_o#.

If #f'(x)>0# on an open interval extending left from #x_o# and #f'(x)<0# on an open interval extending right from #xo#, then #f# has a relative maximum at #x_o#.
If #f'(x)<0# on an open interval extending left from #x_o# and #f'(x)>0# on an open interval extending right from #x_o#, then #f# has a relative minimum at #x_o#.
If #f(x)# has the same sign on both an open interval extending left from #x_# and an open interval extending right from #x_o#, then #f# does not have a relative extremum at #x_o#.

Now we find the first derivative of #f(x)=x/(x^2+3)# hence we have that

#d(f(x))/dx=(x'*(x^2+3)-x(x^2+3)')/(x^2+3)^2=(x^2+3-2x^2)/(x^2+3)^2=(3-x^2)/(x^2+3)#

Hence the first derivative nullifies at #x_o=sqrt3# and #x_o=-sqrt3#

According to the first derivative test f has

  • local maximum at #x=sqrt3# with #f(sqrt3)=1/(2sqrt3)#
  • local minimum at #x=-sqrt3#with #f(-sqrt3)=-1/(2sqrt3)#