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

  • MHB
  • Thread starter hannahSUU
  • Start date
  • Tags
    intervals
In summary, the function f(x)=x^6\ln(x) is concave down on the interval (0, e^{\Large{-\frac{11}{30}}}) and concave up on the interval (e^{\Large{-\frac{11}{30}}}, \infty). The point of inflection occurs at (e^{\Large{-\frac{11}{30}}},-\frac{11}{30}e^{\Large{-\frac{11}{5}}} ).
  • #1
hannahSUU
4
0
\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)
 
Physics news on Phys.org
  • #2
hannahSUU said:
\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?
 
  • #3
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
 

Attachments

  • concavity_01.png
    concavity_01.png
    4.4 KB · Views: 59

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

1. What is the interval of concavity for ${x}^{6}\ln\left({x}\right)$?

The interval of concavity for ${x}^{6}\ln\left({x}\right)$ is from negative infinity to zero and from zero to positive infinity.

2. How do you determine the interval of concavity for a function?

The interval of concavity for a function can be determined by finding the second derivative of the function and setting it equal to zero. The intervals where the second derivative is positive are the intervals of concavity.

3. Why is the interval of concavity for ${x}^{6}\ln\left({x}\right)$ from negative infinity to zero and from zero to positive infinity?

This is because the second derivative of ${x}^{6}\ln\left({x}\right)$ is positive for all values of x except at x=0, where it is undefined. Therefore, the interval of concavity is from negative infinity to zero and from zero to positive infinity.

4. Can the interval of concavity be negative?

No, the interval of concavity cannot be negative. It can only be positive or undefined.

5. How does the interval of concavity affect the graph of a function?

The interval of concavity affects the graph of a function by determining where the graph is concave up or concave down. In the interval where the second derivative is positive, the graph is concave up, and in the interval where the second derivative is negative, the graph is concave down.

Similar threads

Back
Top