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consider the following. (if an answer does not exist, enter dne.) c(x) …

Question

consider the following. (if an answer does not exist, enter dne.)

c(x) = x^{1/5}(x + 6)

(a) find the interval(s) of increase. (enter your answer using interval notation.)

find the interval(s) of decrease. (enter your answer using interval notation.)

(b) find the local minimum value(s). (enter your answers as a comma - separated list.)

find the local maximum value(s). (enter your answers as a comma - separated list.)

(c) find the inflection points.

smaller x - value (x, y) = (

larger x - value (x, y) = (

find the interval(s) where the function is concave up. (enter your answer using interval notation.)

find the interval(s) where the function is concave down. (enter your answer using interval notation.)

Explanation:

Step1: Differentiate \(C(x)\)

Use the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u = x^{\frac{1}{5}}\) and \(v=x + 6\).
\(u^\prime=\frac{1}{5}x^{-\frac{4}{5}}\), \(v^\prime = 1\)
\(C^\prime(x)=\frac{1}{5}x^{-\frac{4}{5}}(x + 6)+x^{\frac{1}{5}}\times1=\frac{x + 6}{5x^{\frac{4}{5}}}+x^{\frac{1}{5}}=\frac{x + 6+5x}{5x^{\frac{4}{5}}}=\frac{6(x + 1)}{5x^{\frac{4}{5}}}\)
Find critical points: \(C^\prime(x)=0\) when \(x=-1\), and \(C^\prime(x)\) 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
  • For \(x>0\), let \(x = 1\), \(C^\prime(1)=\frac{6(1 + 1)}{5(1)^{\frac{4}{5}}}>0\)

Step2: Find second - derivative

\(C^\prime(x)=\frac{6}{5}x^{-\frac{4}{5}}+\frac{6}{5}x^{-\frac{1}{5}}\)
\(C^{\prime\prime}(x)=-\frac{24}{25}x^{-\frac{9}{5}}-\frac{6}{25}x^{-\frac{6}{5}}=-\frac{6(4 + x)}{25x^{\frac{9}{5}}}\)
Find inflection points: \(C^{\prime\prime}(x)=0\) when \(x=-4\), and \(C^{\prime\prime}(x)\) is undefined at \(x = 0\)
Test intervals:

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

Answer:

(a)

  • Interval of increase: \((-1,0)\cup(0,\infty)\)
  • Interval of decrease: \((-\infty,-1)\)

(b)

  • Local minimum value: \(C(-1)=(-1)^{\frac{1}{5}}(-1 + 6)=-5\)
  • Local maximum value: DNE

(c)

  • Smaller \(x\) - value inflection point: When \(x=-4\), \(y=(-4)^{\frac{1}{5}}(-4 + 6)=2(-4)^{\frac{1}{5}}\approx - 2.297\), \((x,y)=(-4,2(-4)^{\frac{1}{5}})\)
  • Larger \(x\) - value inflection point: DNE
  • Concave up interval: \((-\infty,-4)\)
  • Concave down interval: \((-4,0)\cup(0,\infty)\)