consider the equation below. (if an answer does not exist, enter dne.)\n\n$f(x)=x^{7}\\ln (x)$\n\n(a) find…

consider the equation below. (if an answer does not exist, enter dne.)\n\n$f(x)=x^{7}\\ln (x)$\n\n(a) find the interval on which $f$ is increasing. (enter your answer using interval notation.)\n\nfind the interval on which $f$ is decreasing. (enter your answer using interval notation.)\n\n(b) find the local minimum and maximum values of $f$.\n\nlocal minimum value\n\nlocal maximum value\n\n(c) find the inflection point.\n\n$(x,y)=(\\quad)$\n\nfind the interval on which $f$ is concave up. (enter your answer using interval notation.)\n\nfind the interval on which $f$ is concave down. (enter your answer using interval notation.)

consider the equation below. (if an answer does not exist, enter dne.)\n\n$f(x)=x^{7}\\ln (x)$\n\n(a) find the interval on which $f$ is increasing. (enter your answer using interval notation.)\n\nfind the interval on which $f$ is decreasing. (enter your answer using interval notation.)\n\n(b) find the local minimum and maximum values of $f$.\n\nlocal minimum value\n\nlocal maximum value\n\n(c) find the inflection point.\n\n$(x,y)=(\\quad)$\n\nfind the interval on which $f$ is concave up. (enter your answer using interval notation.)\n\nfind the interval on which $f$ is concave down. (enter your answer using interval notation.)

Answer

Explanation:

Step1: Find the first derivative

Use the product rule ((uv)^\prime = u^\prime v+uv^\prime), where (u = x^{7}), (u^\prime=7x^{6}), (v=\ln(x)), (v^\prime=\frac{1}{x}). (f^\prime(x)=7x^{6}\ln(x)+x^{7}\cdot\frac{1}{x}=x^{6}(7\ln(x) + 1))

Step2: Find critical points

Set (f^\prime(x)=0). Since (x^{6}>0) for (x>0) (domain of (y = \ln(x)) is (x>0)), we solve (7\ln(x)+1 = 0). (\ln(x)=-\frac{1}{7}), so (x = e^{-\frac{1}{7}})

  • Test intervals:
    • For (0<x<e^{-\frac{1}{7}}), let (x=\frac{1}{e}). Then (f^\prime(\frac{1}{e})=(\frac{1}{e})^{6}(7\ln(\frac{1}{e})+1)=(\frac{1}{e})^{6}(-7 + 1)<0)
    • For (x>e^{-\frac{1}{7}}), let (x = 1). Then (f^\prime(1)=1^{6}(7\ln(1)+1)=1>0)

Step3: Find the second derivative

Use the product rule again on (f^\prime(x)=x^{6}(7\ln(x)+1)). Let (u = x^{6}), (u^\prime=6x^{5}), (v = 7\ln(x)+1), (v^\prime=\frac{7}{x}) (f^{\prime\prime}(x)=6x^{5}(7\ln(x)+1)+x^{6}\cdot\frac{7}{x}=x^{5}(42\ln(x)+6 + 7)=x^{5}(42\ln(x)+13))

  • Set (f^{\prime\prime}(x)=0). Since (x^{5}>0) for (x>0), solve (42\ln(x)+13 = 0) (\ln(x)=-\frac{13}{42}), so (x = e^{-\frac{13}{42}})
  • Test intervals:
    • For (0<x<e^{-\frac{13}{42}}), let (x=\frac{1}{e}). Then (f^{\prime\prime}(\frac{1}{e})=(\frac{1}{e})^{5}(42\ln(\frac{1}{e})+13)=(\frac{1}{e})^{5}(-42 + 13)<0)
    • For (x>e^{-\frac{13}{42}}), let (x = 1). Then (f^{\prime\prime}(1)=1^{5}(42\ln(1)+13)=13>0)

Step4: Evaluate function at critical points

  • Local minimum: (f(e^{-\frac{1}{7}})=e^{-\frac{7}{7}}\ln(e^{-\frac{1}{7}})=-\frac{1}{7e})
  • Inflection - point (y) - value: (f(e^{-\frac{13}{42}})=e^{-\frac{13}{6}}\ln(e^{-\frac{13}{42}})=-\frac{13}{42}e^{-\frac{13}{6}})

Answer:

(a)

  • Increasing: ((e^{-\frac{1}{7}},\infty))
  • Decreasing: ((0,e^{-\frac{1}{7}})) (b)
  • Local minimum value: (-\frac{1}{7e})
  • Local maximum value: DNE (c)
  • Inflection point: ((e^{-\frac{13}{42}},-\frac{13}{42}e^{-\frac{13}{6}}))
  • Concave up: ((e^{-\frac{13}{42}},\infty))
  • Concave down: ((0,e^{-\frac{13}{42}}))