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) (1 points) find the domain of $f$.

domain of $f$: $(-\infty, -1)(-1, 1)(1, \infty)$
(b) (6 points) set up a chart to determine where $f$ is concave up (c.u.) and concave down (c.d.)

(c) (2 points) $f$ c.u.:________ $f$ c.d.:________
(d) (1 points) find all inflection points on the graph of $f$.

inflection point(s):__________

Explanation:

Step1: Find domain of f(x)

Denominator \(x^3 - 1 = 0 \Rightarrow x = 1\). So domain excludes \(x=1\).

Step2: Find critical points for concavity

Second derivative \(f''(x) = \frac{6x^2(x^3 + 2)}{(x^3 - 1)^3}\). Set numerator/denominator to 0: \(x=0\), \(x=\sqrt[3]{-2}\approx-1.26\), \(x=1\). Divide domain into intervals: \((-\infty, -\sqrt[3]{2})\), \((-\sqrt[3]{2}, 0)\), \((0,1)\), \((1, \infty)\).

Step3: Test concavity in intervals

• \((-\infty, -\sqrt[3]{2})\): Pick \(x=-2\), \(f''(-2)=\frac{6(4)(-8+2)}{(-8-1)^3}=\frac{24(-6)}{-729}>0\) (c.u.)
• \((-\sqrt[3]{2},0)\): Pick \(x=-1\), \(f''(-1)=\frac{6(1)(-1+2)}{(-1-1)^3}=\frac{6(1)}{-8}<0\) (c.d.)
• \((0,1)\): Pick \(x=0.5\), \(f''(0.5)=\frac{6(0.25)(0.125+2)}{(0.125-1)^3}=\frac{1.5(2.125)}{(-0.659)^3}<0\) (c.d.)
• \((1,\infty)\): Pick \(x=2\), \(f''(2)=\frac{6(4)(8+2)}{(8-1)^3}=\frac{24(10)}{343}>0\) (c.u.)

Step4: Find inflection points

Check where \(f''(x)=0\): \(x=0\) or \(x=-\sqrt[3]{2}\). At \(x=0\), \(f(0)=0\); at \(x=-\sqrt[3]{2}\), \(f(-\sqrt[3]{2})=\frac{-\sqrt[3]{2}}{(-2)-1}=\frac{\sqrt[3]{2}}{3}\). Concavity changes at both points.

Answer:

(a) \((-\infty,1)\cup(1,\infty)\)
(b)

Interval	Sign of \(f''(x)\)	Concavity
\((-\sqrt[3]{2},0)\)	Negative	c.d.
\((0,1)\)	Negative	c.d.
\((1,\infty)\)	Positive	c.u.

(c) \(f\) c.u.: \((-\infty,-\sqrt[3]{2})\cup(1,\infty)\); \(f\) c.d.: \((-\sqrt[3]{2},1)\)
(d) \((-\sqrt[3]{2},\frac{\sqrt[3]{2}}{3})\) and \((0,0)\)