How do you find the local maximum and minimum values of #f(x) = x / (x^2 + 81)# using both the First and Second Derivative Tests?

1 Answer
Jan 2, 2018

Minimum #(-9,-1/18)# , Maximum #(9,1/18)#

Explanation:

#f(x)=x/(x^2+81)#

#D_f=RR#

  • #f'(x)=(x^2+81-x(2x))/(x^2+81)^2# #=#

#(-x^2+81)/(x^2+81)^2# #=#

#-(x^2-81)/(x^2+81)^2#

#f'(x)=0 ##<=> x^2-81=0 <=> x^2=81 <=> (x=9 or x=-9)#

Let's plug in #f'# a value , #x_1in##(-9,9)# . For example, for #x_1=0#
we get

#f'(0)=-(-81)/81^2=81/81^2>0#

Let's plug in #f'# a value which is #>9# . For example, for #x_2=10#
we get

#f'(10)=-(100-81)/(100+81)^2=-19/181^2<0#

Let's plug in #f'# a value which is #<-9# . For example, for #x_3=-10#
we get

#f'(-10)=-(100-81)/(100+81)^2=-19/181^2<0#

As a result we have:

  • #f# continuous in #(-oo,-9]# and #f'(x)<0# for #x##in##(-oo,-9)#
    so #f# is strictly decreasing in #(-oo,-9]#

  • #f# continuous in #[-9,9]# and #f'(x)>0# for #x##in##(-9,-9)#
    so #f# is strictly increasing in #[-9,9]#

  • #f# continuous in #[9,+oo)# and #f'(x)<0# for #x##in##(9,+oo)#
    so #f# is strictly decreasing in #[9,+oo)#

#f# is decreasing in #(-oo,-9]# and increasing in #[-9,9]#
therefore #f# has a local minimum at #x=-9#

#f# is increasing in #[-9,9]# and decreasing in #[9,+oo)#
therefore #f# has a local maximum at #x=9#

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