Question

consider the function

$f(x)=x^{3}\ln x,\\ x > 0$.

a) determine the intervals on which $f$ is concave up and concave down.

$f$ is concave up on:

$f$ is concave down on:

b) based on your answer to part (a), determine the inflection points of $f$. each point should be entered as an ordered pair (that is, in the form $(x,y)$).

(separate multiple answers by commas.)

(round to four decimal places as needed.)

note: when using interval notation in webwork, remember that:

you use inf for $\infty$ and -inf for $-\infty$,

and use u for the union symbol.

enter dne if an answer does not exist.

partial credit on this problem.

Explanation:

Step1: Find the first - derivative

Using the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u = x^{3}\) and \(v=\ln x\).
\(u^\prime=3x^{2}\), \(v^\prime=\frac{1}{x}\)
\(f^\prime(x)=3x^{2}\ln x + x^{3}\cdot\frac{1}{x}=3x^{2}\ln x+x^{2}=x^{2}(3\ln x + 1)\)

Step2: Find the second - derivative

Using the product rule again. Let \(u = x^{2}\) and \(v = 3\ln x+1\)
\(u^\prime = 2x\), \(v^\prime=\frac{3}{x}\)
\(f^{\prime\prime}(x)=2x(3\ln x + 1)+x^{2}\cdot\frac{3}{x}=6x\ln x+2x + 3x=6x\ln x+5x=x(6\ln x + 5)\)

Step3: Find the critical points of \(f^{\prime\prime}(x)\)

Set \(f^{\prime\prime}(x)=0\), since \(x>0\), we solve \(6\ln x+5 = 0\)
\(\ln x=-\frac{5}{6}\), then \(x = e^{-\frac{5}{6}}\approx0.4346\)

Step4: Test the intervals

• For the interval \((0,e^{-\frac{5}{6}})\), let \(x = 0.1\)

\(f^{\prime\prime}(0.1)=0.1\times(6\ln(0.1)+5)=0.1\times(6\times(- 2.3026)+5)=0.1\times(-13.8156 + 5)=-0.8816<0\)

• For the interval \((e^{-\frac{5}{6}},+\infty)\), let \(x = 1\)

\(f^{\prime\prime}(1)=1\times(6\ln(1)+5)=5>0\)

Answer:

a) \(f\) is concave up on: \((e^{-\frac{5}{6}},+\infty)\) (or \((0.4346,+\infty)\))
\(f\) is concave down on: \((0,e^{-\frac{5}{6}})\) (or \((0,0.4346)\))
b) Inflection point: \((e^{-\frac{5}{6}},e^{-\frac{5}{6}}^{3}\ln(e^{-\frac{5}{6}}))\)
Calculate \(y\) - coordinate:
\(y = e^{-\frac{5}{2}}\times(-\frac{5}{6})\approx0.4346^{3}\times(-\frac{5}{6})\approx0.0821\times(-0.8333)\approx - 0.0684\)
Inflection point: \((0.4346,-0.0684)\)