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

1 Answer
Jul 23, 2018

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

Explanation:

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.