Question

f(x)=3x^{\frac{1}{3}} - x
furthest left local extrema: (□,□)
furthest right local extrema: (□,□)
increasing on the interval: □
decreasing on the interval: □
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Explanation:

Step1: Find the derivative

Use the power - rule. If $y = ax^n$, then $y'=nax^{n - 1}$. For $f(x)=3x^{\frac{1}{3}}-x$, $f'(x)=3\times\frac{1}{3}x^{-\frac{2}{3}}-1=x^{-\frac{2}{3}}-1=\frac{1}{x^{\frac{2}{3}}}-1=\frac{1 - x^{\frac{2}{3}}}{x^{\frac{2}{3}}}=\frac{(1 - x^{\frac{1}{3}})(1 + x^{\frac{1}{3}})}{x^{\frac{2}{3}}}$.

Step2: Find critical points

Set $f'(x)=0$. Then $\frac{(1 - x^{\frac{1}{3}})(1 + x^{\frac{1}{3}})}{x^{\frac{2}{3}}}=0$. The numerator $(1 - x^{\frac{1}{3}})(1 + x^{\frac{1}{3}})=0$ gives $x = 1$ or $x=-1$. The derivative is undefined at $x = 0$, but $f(x)$ is defined at $x = 0$.

Step3: Test intervals for increasing and decreasing

Choose test points in the intervals $(-\infty,-1)$, $(-1,0)$, $(0,1)$ and $(1,\infty)$.

• For $x=-8$ in $(-\infty,-1)$, $f'(-8)=\frac{1-(-8)^{\frac{2}{3}}}{(-8)^{\frac{2}{3}}}=\frac{1 - 4}{4}=-\frac{3}{4}<0$, so $f(x)$ is decreasing on $(-\infty,-1)$.
• For $x =-\frac{1}{8}$ in $(-1,0)$, $f'(-\frac{1}{8})=\frac{1-(-\frac{1}{8})^{\frac{2}{3}}}{(-\frac{1}{8})^{\frac{2}{3}}}=\frac{1-\frac{1}{4}}{\frac{1}{4}} = 3>0$, so $f(x)$ is increasing on $(-1,0)$.
• For $x=\frac{1}{8}$ in $(0,1)$, $f'(\frac{1}{8})=\frac{1-(\frac{1}{8})^{\frac{2}{3}}}{(\frac{1}{8})^{\frac{2}{3}}}=\frac{1-\frac{1}{4}}{\frac{1}{4}}=3>0$, so $f(x)$ is increasing on $(0,1)$.
• For $x = 8$ in $(1,\infty)$, $f'(8)=\frac{1-8^{\frac{2}{3}}}{8^{\frac{2}{3}}}=\frac{1 - 4}{4}=-\frac{3}{4}<0$, so $f(x)$ is decreasing on $(1,\infty)$.

Step4: Find local extrema

Since $f(x)$ changes from decreasing to increasing at $x=-1$, $f(-1)=3(-1)^{\frac{1}{3}}-(-1)=-3 + 1=-2$ is a local minimum. Since $f(x)$ changes from increasing to decreasing at $x = 1$, $f(1)=3(1)^{\frac{1}{3}}-1=3 - 1=2$ is a local maximum.

Answer:

Furthest left local extrema: $(-1,-2)$
Furthest right local extrema: $(1,2)$
Increasing on the interval: $(-1,1)$
Decreasing on the interval: $(-\infty,-1)\cup(1,\infty)$