consider the following. (if an answer does not exist, enter dne.)\n\n$f(x)=x^{2 / 3}(x - 6)$\n\n(a) find the…

consider the following. (if an answer does not exist, enter dne.)\n\n$f(x)=x^{2 / 3}(x - 6)$\n\n(a) find the interval(s) on which $f$ is increasing. (enter your answer using interval notation.)\n\n(b) find the interval(s) on which $f$ is decreasing. (enter your answer using interval notation.)\n\n(c) find the local minimum and maximum value of $f$. (round your answer to two decimal places.)\n\nlocal minimum value\n\nlocal maximum value

consider the following. (if an answer does not exist, enter dne.)\n\n$f(x)=x^{2 / 3}(x - 6)$\n\n(a) find the interval(s) on which $f$ is increasing. (enter your answer using interval notation.)\n\n(b) find the interval(s) on which $f$ is decreasing. (enter your answer using interval notation.)\n\n(c) find the local minimum and maximum value of $f$. (round your answer to two decimal places.)\n\nlocal minimum value\n\nlocal maximum value

Answer

Explanation:

Step1: Find the derivative of (f(x))

Using the product rule ((uv)^\prime = u^\prime v+uv^\prime), where (u = x^{\frac{2}{3}}) and (v=x - 6). (u^\prime=\frac{2}{3}x^{-\frac{1}{3}}) and (v^\prime = 1) (f^\prime(x)=\frac{2}{3}x^{-\frac{1}{3}}(x - 6)+x^{\frac{2}{3}}\times1=\frac{2(x - 6)}{3x^{\frac{1}{3}}}+x^{\frac{2}{3}}=\frac{2x-12 + 3x}{3x^{\frac{1}{3}}}=\frac{5x-12}{3x^{\frac{1}{3}}})

Step2: Find the critical points

Set (f^\prime(x)=0), then (\frac{5x - 12}{3x^{\frac{1}{3}}}=0), which gives (5x-12 = 0), so (x=\frac{12}{5}=2.4) Also, (f^\prime(x)) is undefined at (x = 0) (since the denominator (3x^{\frac{1}{3}}=0) when (x = 0))

Step3: Test the intervals

  • For (x<0), let (x=-1), (f^\prime(-1)=\frac{5\times(-1)-12}{3\times(-1)^{\frac{1}{3}}}=\frac{-17}{-3}=\frac{17}{3}>0)
  • For (0<x<2.4), let (x = 1), (f^\prime(1)=\frac{5\times1-12}{3\times1^{\frac{1}{3}}}=\frac{-7}{3}<0)
  • For (x>2.4), let (x=3), (f^\prime(3)=\frac{5\times3-12}{3\times3^{\frac{1}{3}}}=\frac{3}{3\times3^{\frac{1}{3}}}=\frac{1}{3^{\frac{1}{3}}}>0)

Step4: Answer part (a)

The function (f(x)) is increasing when (f^\prime(x)>0). The intervals are ((-\infty,0)\cup(\frac{12}{5},\infty))

Step5: Answer part (b)

The function (f(x)) is decreasing when (f^\prime(x)<0). The interval is ((0,\frac{12}{5}))

Step6: Find the local extrema

  • (f(0)=0^{\frac{2}{3}}(0 - 6)=0)
  • (f(\frac{12}{5})=(\frac{12}{5})^{\frac{2}{3}}(\frac{12}{5}-6)=(\frac{12}{5})^{\frac{2}{3}}\times(-\frac{18}{5})\approx-5.70)

Answer:

(a) ((-\infty,0)\cup(\frac{12}{5},\infty)) (b) ((0,\frac{12}{5})) (c) local minimum value: (-5.70), local maximum value: (0.00)