consider the following. (if an answer does not exist, enter dne.)\n\nc(x) = x^{1/3}(x + 4)\n\n(a) find the…

Explanation:

Step1: Find the first derivative

Use the product rule ((uv)^\prime = u^\prime v+uv^\prime), where (u = x^{1/3}), (u^\prime=\frac{1}{3}x^{-2/3}), and (v=x + 4), (v^\prime = 1). [ \begin{align*} C^\prime(x)&=\frac{1}{3}x^{-2/3}(x + 4)+x^{1/3}\times1\ &=\frac{x + 4}{3x^{2/3}}+x^{1/3}\ &=\frac{x + 4+3x}{3x^{2/3}}\ &=\frac{4x + 4}{3x^{2/3}}\ &=\frac{4(x + 1)}{3x^{2/3}} \end{align*} ] Set (C^\prime(x)=0), then (x=- 1). The domain of (C(x)) is ((-\infty,\infty)) (since (y = x^{1/3}) is defined for all real (x)). Test intervals:

• For (x<-1), let (x=-2), (C^\prime(-2)=\frac{4(-2 + 1)}{3(-2)^{2/3}}=\frac{-4}{3\times2^{2/3}}<0)
• For (-1<x<0), let (x=-\frac{1}{2}), (C^\prime(-\frac{1}{2})=\frac{4(-\frac{1}{2}+1)}{3(-\frac{1}{2})^{2/3}}=\frac{2}{3\times(\frac{1}{2})^{2/3}}>0)
• For (x>0), let (x = 1), (C^\prime(1)=\frac{4(1 + 1)}{3\times1^{2/3}}=\frac{8}{3}>0)

Step2: Find the second derivative

Use the quotient rule ((\frac{u}{v})^\prime=\frac{u^\prime v-uv^\prime}{v^{2}}), where (u = 4x + 4), (u^\prime=4), and (v = 3x^{2/3}), (v^\prime = 2x^{-1/3}) [ \begin{align*} C^{\prime\prime}(x)&=\frac{4\times3x^{2/3}-(4x + 4)\times2x^{-1/3}}{9x^{4/3}}\ &=\frac{12x^{2/3}-8x^{2/3}-8x^{-1/3}}{9x^{4/3}}\ &=\frac{4x^{2/3}-8x^{-1/3}}{9x^{4/3}}\ &=\frac{4x - 8}{9x^{5/3}} \end{align*} ] Set (C^{\prime\prime}(x)=0), then (4x-8=0), (x = 2). Test intervals:

• For (x<0), let (x=-1), (C^{\prime\prime}(-1)=\frac{4(-1)-8}{9(-1)^{5/3}}=\frac{-12}{-9}=\frac{4}{3}>0)
• For (0<x<2), let (x = 1), (C^{\prime\prime}(1)=\frac{4\times1-8}{9\times1^{5/3}}=\frac{-4}{9}<0)
• For (x>2), let (x = 3), (C^{\prime\prime}(3)=\frac{4\times3-8}{9\times3^{5/3}}=\frac{4}{9\times3^{5/3}}>0)

Step3: Analyze results for part (a)

The function (C(x)) is increasing when (C^\prime(x)>0). From the first - derivative test, (C(x)) is increasing on ((-1,0)\cup(0,\infty)) (since (C^\prime(x)>0) for (-1 < x<0) and (x>0)). The function (C(x)) is decreasing when (C^\prime(x)<0). From the first - derivative test, (C(x)) is decreasing on ((-\infty,-1))

Step4: Analyze results for part (b)

Since (C(x)) changes from decreasing ((C^\prime(x)<0)) to increasing ((C^\prime(x)>0)) at (x=-1), the local minimum value is (C(-1)=(-1)^{1/3}(-1 + 4)=-3) There is no local maximum value (because the function only changes from decreasing to increasing at (x=-1) and the derivative is positive for (x>-1,x\neq0))

Step5: Analyze results for part (c)

When (x = 0), (C^{\prime\prime}(x)) is undefined. When (x = 2), (C(2)=2^{1/3}(2 + 4)=6\times2^{1/3}) The inflection points:

• For (x = 0), (C(0)=0), but (C^{\prime\prime}(x)) changes sign around (x = 0) (from positive when (x<0) to negative when (0<x<2))
• For (x = 2), (C(2)=6\times2^{1/3}) The function (C(x)) is concave up when (C^{\prime\prime}(x)>0), which is on ((-\infty,0)\cup(2,\infty)) The function (C(x)) is concave down when (C^{\prime\prime}(x)<0), which is on ((0,2))

Answer:

(a)

• Interval of increase: ((-1,0)\cup(0,\infty))
• Interval of decrease: ((-\infty,-1)) (b)
• Local minimum value: (-3)
• Local maximum value: DNE (c)
• Smaller (x) - value inflection point: ((0,0))
• Larger (x) - value inflection point: ((2,6\sqrt[3]{2}))
• Concave up interval: ((-\infty,0)\cup(2,\infty))
• Concave down interval: ((0,2))