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
  • 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