Interval of Concavity for ${x}^{6}\ln\left({x}\right)$

[hannahSUU]

\begin{equation} {x}^{6}\ln\left({x}\right)\end{equation}

Find the interval where the function is concave up and down.

The answers I got are:
Concave up: (ln(2), infinity)
Concave down: (0, ln(2))

and for the life of me I cannot figure out why they aren't correct.

any help? (heart)

 

[MarkFL]

\begin{equation} {x}^{6}\ln\left({x}\right)\end{equation}

Find the interval where the function is concave up and down.

The answers I got are:
Concave up: (ln(2), infinity)
Concave down: (0, ln(2))

and for the life of me I cannot figure out why they aren't correct.

any help? (heart)

Can you show your working in computing the second derivative?

 

[MarkFL]

As a follow-up, we are given:

\(\displaystyle f(x)=x^6\ln(x)\)

The first thing I would observe is that the domain of this function is $0<x$.

Using the product rule, we find:

\(\displaystyle f'(x)=\left(6x^5\right)\left(\ln(x)\right)+\left(x^6\right)\left(\frac{1}{x}\right)=x^5\left(6\ln(x)+1\right)\)

Using the product rule again, we find:

\(\displaystyle f''(x)=\left(5x^4\right)\left(6\ln(x)+1\right)+\left(x^5\right)\left(\frac{6}{x}\right)=x^4\left(30\ln(x)+11\right)\)

As $0<x$, we need only look at the critical value that comes from:

\(\displaystyle 30\ln(x)+11=0\implies x=e^{\Large{-\frac{11}{30}}}\)

Using test values on either side of this critical value, such as $e^{-1}$ or $e$, we see that we have:

\(\displaystyle 30\ln(x)+11<0\) on $\left(0,e^{\Large{-\frac{11}{30}}}\right)$ and thus $f$ is concave down on this interval.

\(\displaystyle 30\ln(x)+11>0\) on $\left(e^{\Large{-\frac{11}{30}}},\infty\right)$ and thus $f$ is concave up on this interval.

Thus, we have a point of inflection at:

\(\displaystyle \left(e^{\Large{-\frac{11}{30}}},f\left(e^{\Large{-\frac{11}{30}}}\right)\right)=\left(e^{\Large{-\frac{11}{30}}},-\frac{11}{30}e^{\Large{-\frac{11}{5}}}\right)\)

Here is a plot of the given function with the point of inflection shown:

View attachment 7538