c(x)=x^{1 / 5}(x + 6)\n(a) find the interval(s) of increase. (enter your answer using interval…

c(x)=x^{1 / 5}(x + 6)\n(a) find the interval(s) of increase. (enter your answer using interval notation.)\n(-1,0),(0,\\infty)\nfind the interval(s) of decrease. (enter your answer using interval notation.)\n(-\\infty, 1)\n(b) find the local minimum value(s). (enter your answers as a comma - separated list.)\n-5\nfind the local maximum value(s). (enter your answers as a comma - separated list.)\ndne\n(c) find the inflection points.\nsmaller x - value (x,y)=( - 2.297)\nlarger x - value (x,y)=(dne)\nfind the interval(s) where the function is concave up. (enter your answer using interval notation.)\n(-\\infty,-4)\nfind the interval(s) where the function is concave down. (enter your answer using interval notation.)\n(-4,0),(0,\\infty)

c(x)=x^{1 / 5}(x + 6)\n(a) find the interval(s) of increase. (enter your answer using interval notation.)\n(-1,0),(0,\\infty)\nfind the interval(s) of decrease. (enter your answer using interval notation.)\n(-\\infty, 1)\n(b) find the local minimum value(s). (enter your answers as a comma - separated list.)\n-5\nfind the local maximum value(s). (enter your answers as a comma - separated list.)\ndne\n(c) find the inflection points.\nsmaller x - value (x,y)=( - 2.297)\nlarger x - value (x,y)=(dne)\nfind the interval(s) where the function is concave up. (enter your answer using interval notation.)\n(-\\infty,-4)\nfind the interval(s) where the function is concave down. (enter your answer using interval notation.)\n(-4,0),(0,\\infty)

Answer

Explanation:

Step1: Find the first - derivative

Using the product rule ((uv)^\prime = u^\prime v+uv^\prime), where (u = x^{\frac{1}{5}}) and (v=x + 6). (u^\prime=\frac{1}{5}x^{-\frac{4}{5}}) and (v^\prime = 1). So (C^\prime(x)=\frac{1}{5}x^{-\frac{4}{5}}(x + 6)+x^{\frac{1}{5}}\times1=\frac{x + 6+5x}{5x^{\frac{4}{5}}}=\frac{6(x + 1)}{5x^{\frac{4}{5}}}). Set (C^\prime(x)=0), then (x=-1). The function (C^\prime(x)) is undefined at (x = 0). Test intervals:

  • For (x<-1), let (x=-2), (C^\prime(-2)=\frac{6(-2 + 1)}{5(-2)^{\frac{4}{5}}}<0).
  • For (-1<x<0), let (x=-\frac{1}{2}), (C^\prime(-\frac{1}{2})=\frac{6(-\frac{1}{2}+1)}{5(-\frac{1}{2})^{\frac{4}{5}}}>0).
  • For (x>0), let (x = 1), (C^\prime(1)=\frac{6(1 + 1)}{5(1)^{\frac{4}{5}}}>0).

Step2: Find the second - derivative

Using the quotient rule ((\frac{u}{v})^\prime=\frac{u^\prime v-uv^\prime}{v^{2}}), where (u = 6(x + 1)) and (v = 5x^{\frac{4}{5}}). (u^\prime=6) and (v^\prime=4x^{-\frac{1}{5}}). (C^{\prime\prime}(x)=\frac{6\times5x^{\frac{4}{5}}-6(x + 1)\times4x^{-\frac{1}{5}}}{25x^{\frac{8}{5}}}=\frac{30x-24(x + 1)}{25x^{\frac{9}{5}}}=\frac{6(x-4)}{25x^{\frac{9}{5}}}). Set (C^{\prime\prime}(x)=0), then (x = 4). The function (C^{\prime\prime}(x)) is undefined at (x = 0). Test intervals:

  • For (x<0), let (x=-1), (C^{\prime\prime}(-1)=\frac{6(-1-4)}{25(-1)^{\frac{9}{5}}}>0).
  • For (0<x<4), let (x = 1), (C^{\prime\prime}(1)=\frac{6(1 - 4)}{25(1)^{\frac{9}{5}}}<0).
  • For (x>4), let (x = 5), (C^{\prime\prime}(5)=\frac{6(5 - 4)}{25(5)^{\frac{9}{5}}}>0).

Answer:

(a) Interval of increase: ((-1,0)\cup(0,\infty)) Interval of decrease: ((-\infty,-1)) (b) Local minimum value: (-5) Local maximum value: DNE (c) Inflection points: ((4,\ 4^{\frac{1}{5}}(4 + 6))=(4,10\times4^{\frac{1}{5}})\approx(4,16.79)) Interval where the function is concave up: ((-\infty,0)\cup(4,\infty)) Interval where the function is concave down: ((0,4))