On what intervals the following equation is concave up, concave down and where it's inflection point is (x,y) #f(x)=x^8(ln(x))#?

1 Answer

Answer:

  • if #0 < x < e^(-15/56)# then #f# is concave down;
  • if #x > e^(-15/56)# then #f# is concave up;
  • #x=e^(-15/56)# is a (falling) inflection point

Explanation:

To analyze concavity and inflection points of a twice differentiable function #f#, we can study the positivity of the second derivative. In fact, if #x_0# is a point in the domain of #f#, then:

  • if #f''(x_0)>0#, then #f# is concave up in a neighborhood of #x_0#;
  • if #f''(x_0)<0#, then #f# is concave down in a neighborhood of #x_0#;
  • if #f''(x_0)=0# and the sign of #f''# on a sufficiently small right-neighborhood of #x_0# is opposite to the sign of #f''# on a sufficiently small left-neighborhood of #x_0#, then #x=x_0# is called an inflection point of #f#.

In the specific case of #f(x)=x^8 ln(x)#, we have a function whose domain has to be restricted to the positive reals #RR^+#.
The first derivative is
#f'(x)=8x^7 ln(x)+x^8 1/x=x^7[8 ln(x)+1]#
The second derivative is
#f''(x)=7x^6[8 ln(x) +1]+x^7 8/x=x^6[56ln(x)+15]#

Let's study the positivity of #f''(x)#:

  • #x^6>0 iff x ne 0#
  • #56ln(x)+15>0 iff ln(x)> -15/56 iff x>e^(-15/56)#

So, considering that the domain is #RR^+#, we get that

  • if #0 < x < e^(-15/56)# then #f''(x)<0# and #f# is concave down;
  • if #x > e^(-15/56)# then #f''(x)>0# and #f# is concave up;
  • if #x=e^(-15/56)# then #f''(x)=0#. Considering that on the left of this point #f''# is negative and on the right it is positive, we conclude that #x=e^(-15/56)# is a (falling) inflection point