Question

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

$f(x)=x^{2 / 3}(x - 6)$

(a) find the interval(s) on which $f$ is increasing. (enter your answer using interval notation.)

(b) find the interval(s) on which $f$ is decreasing. (enter your answer using interval notation.)

(c) find the local minimum and maximum value of $f$. (round your answer to two decimal places.)

local minimum value

local maximum value

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
• 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: \(0.00\)