consider the function\n\n$f(x)=x^{3}\\ln x,\\ x > 0$.\n\na) determine the intervals on which $f$ is concave…

consider the function\n\n$f(x)=x^{3}\\ln x,\\ x > 0$.\n\na) determine the intervals on which $f$ is concave up and concave down.\n\n$f$ is concave up on:\n\n$f$ is concave down on:\n\nb) 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 $(x,y)$).\n\n(separate multiple answers by commas.)\n\n(round to four decimal places as needed.)\n\nnote: when using interval notation in webwork, remember that:\n\n you use inf for $\\infty$ and -inf for $-\\infty$,\n\n and use u for the union symbol.\n\n enter dne if an answer does not exist.\n\n partial credit on this problem.

consider the function\n\n$f(x)=x^{3}\\ln x,\\ x > 0$.\n\na) determine the intervals on which $f$ is concave up and concave down.\n\n$f$ is concave up on:\n\n$f$ is concave down on:\n\nb) 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 $(x,y)$).\n\n(separate multiple answers by commas.)\n\n(round to four decimal places as needed.)\n\nnote: when using interval notation in webwork, remember that:\n\n you use inf for $\\infty$ and -inf for $-\\infty$,\n\n and use u for the union symbol.\n\n enter dne if an answer does not exist.\n\n partial credit on this problem.

Answer

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))