How do you find the inflection point and concavity of #f(x)=x^(4)*ln (x)#?

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Answer 1 · 2018-07-23T14:02:47

#f_min = f( e^(-1/4)) approx -0.092#.
#f(x)# is concave up.

#f(x) = x^4ln(x)#

Since #lnx# is defined for #x>0 -> f(x)# is defined for #x >0#

To find an inflection point we need to find where #f'(x)=0#

#f'(x) = 4x^3ln(x) + x^4*1/x# [Chain rule]

#= 4x^3ln(x) + x^3#

#f'(x)=0 -> 4x^3ln(x) + x^3 =0#

#=>x^3(4lnx +1) =0#

#:. x=0 or (4lnx +1)=0#

Since #x>0# consider #4lnx = -1#

#=>lnx = -1/4#

#x= e^(-1/4)#

#x approx 0.7788#

Now, let's look at the graph of #f(x)# below.

graph{ x^4ln(x) [-0.234, 1.452, -0.2815, 0.561]}

We can see that #f(e^(-1/4))# is a minimum value of #f(x) approx -0.092#.
#f(x)# is concave up at this point.

From the nature of #f(x)# we may conclude that this is the only inflection point.