Question

1. given the function

$f(x) = \frac{x}{x^3 - 1}$
with first derivative $f(x) = \frac{-2x^3 - 1}{(x^3 - 1)^2}$
and second derivative $f(x) = \frac{6x^2(x^3 + 2)}{(x^3 - 1)^3}$
(a) (2 points) find the domain of $f$ and the domain of $f$.
domain of $f$: $(-\infty, 1) \cup (1, \infty)$ domain of $f$: $(-\infty, 1) \cup (1, \infty)$
(b) (4 points)find the critical values of the function $f$.
(c) (3 points) set up a chart to determine where $f$ is increasing and
decreasing.
(d) (2 points) $f$ increasing: __________ $f$ decreasing: __________
(e) (3 points) find all relative extrema points on the graph of $f$. give
answers accurate to one decimal place.
relative max: __________ relative min: __________

Explanation:

Step1: Find critical values

Critical values occur where \(f^{\prime}(x) = 0\) or \(f^{\prime}(x)\) is undefined.
\(f^{\prime}(x)=\frac{-2x^{3}-1}{(x^{3}-1)^{2}}\)
Set numerator equal to zero: \(-2x^{3}-1 = 0\)
\(2x^{3}=-1\)
\(x^{3}=-\frac{1}{2}\)
\(x =-\sqrt[3]{\frac{1}{2}}\approx - 0.8\)
\(f^{\prime}(x)\) is undefined at \(x = 1\) (since denominator \((x^{3}-1)^{2}=0\) when \(x = 1\)), but \(x = 1\) is not in the domain of \(f(x)\) (from part (a), domain of \(f(x)\) is \((-\infty,1)\cup(1,\infty)\))

Step2: Set up intervals for increasing/decreasing

Intervals are \((-\infty,-\sqrt[3]{\frac{1}{2}})\) and \((-\sqrt[3]{\frac{1}{2}},1)\cup(1,\infty)\)
Take a test point in \((-\infty,-\sqrt[3]{\frac{1}{2}})\), say \(x=-1\)
\(f^{\prime}(-1)=\frac{-2(-1)^{3}-1}{((-1)^{3}-1)^{2}}=\frac{2 - 1}{(-2)^{2}}=\frac{1}{4}>0\)
Take a test point in \((-\sqrt[3]{\frac{1}{2}},1)\cup(1,\infty)\), say \(x = 0\)
\(f^{\prime}(0)=\frac{-2(0)^{3}-1}{(0^{3}-1)^{2}}=-1<0\)

Step3: Determine increasing/decreasing

Since \(f^{\prime}(x)>0\) on \((-\infty,-\sqrt[3]{\frac{1}{2}})\), \(f(x)\) is increasing on \((-\infty,-\sqrt[3]{\frac{1}{2}})\)
Since \(f^{\prime}(x)<0\) on \((-\sqrt[3]{\frac{1}{2}},1)\cup(1,\infty)\), \(f(x)\) is decreasing on \((-\sqrt[3]{\frac{1}{2}},1)\cup(1,\infty)\)

Step4: Find relative extrema

Since \(f(x)\) changes from increasing to decreasing at \(x =-\sqrt[3]{\frac{1}{2}}\)
\(f(-\sqrt[3]{\frac{1}{2}})=\frac{-\sqrt[3]{\frac{1}{2}}}{(-\sqrt[3]{\frac{1}{2}})^{3}-1}=\frac{-\sqrt[3]{\frac{1}{2}}}{-\frac{1}{2}-1}=\frac{-\sqrt[3]{\frac{1}{2}}}{-\frac{3}{2}}=\frac{2}{3}\sqrt[3]{\frac{1}{2}}\approx0.3\)

Answer:

(b) Critical value: \(x\approx - 0.8\)
(d) \(f\) increasing: \((-\infty,-\sqrt[3]{\frac{1}{2}})\); \(f\) decreasing: \((-\sqrt[3]{\frac{1}{2}},1)\cup(1,\infty)\)
(e) Relative max: \((-0.8,0.3)\)