What intervals is this function: #f(x)= x^5(lnx)# increasing and decreasing?

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Answer 1 · 2017-12-29T19:25:57

#f# strictly decreasing in #(0,e^(-1/5)]# & #f# strictly increasing in #[e^(-1/5),+oo)#

#f(x)=x^5lnx# , #D_f=(0,+oo)#

For #x##in##D_f#,
#f'(x)=(x^5lnx)' =#

#(x^5)'lnx+x^5(lnx)'# #=#

#5x^4lnx+x^5/x# #=#

#5x^4lnx+x^4# #=#

#x^4(5lnx+1)#

#5lnx+1=0# #<=># #5lnx=-1# #<=># #lnx=-1/5# #<=># #x=e^(-1/5)# #~=# #0.81#

For #x=1/2=0.5# ---> #f'(1/2)=5*1/2^4ln(1/2)+1/2^4# #=#
#5/16(ln1-ln2)+1/16# #=# #-ln2*5/16+1/16# #<0#

because #ln2~=0.7#
#1/16=0.0625#

For #x=1# ----> #f'(1)=1>0#

As a result we have,

#f# continuous in #(0,e^(-1/5)]# with #f'(x)<0# when #x##in##(0,e^(-1/5))# so #f# strictly decreasing in #(0,e^(-1/5)]#
#f# continuous in #[e^(-1/5),+oo)# with #f'(x)>0# when #x##in##(e^(-1/5),+oo)# so #f# strictly increasing in #[e^(-1/5),+oo)#

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