Question

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))`?

Answer 1

• 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