Topics Covered:
Information from the First Derivative \(f'(x)\):
Information from the Second Derivative \(f''(x)\):
Additional Important Concepts:
Find the first derivative:
\[f'(x) = x^2 - 16\]Find critical points by setting \(f'(x) = 0\):
\[x^2 - 16 = 0\] \[x^2 = 16\] \[x = 4 \text{ and } x = -4\]Create a sign chart for \(f'(x)\):
| Interval | Test Point | Sign of \(x^2 - 16\) | Sign of \(f'(x)\) | Behavior |
|---|---|---|---|---|
| \((-\infty, -4)\) | \(x = -5\) | \(25 - 16 > 0\) | \(+\) | Increasing ↑ |
| \((-4, 4)\) | \(x = 0\) | \(0 - 16 < 0\) | \(-\) | Decreasing ↓ |
| \((4, \infty)\) | \(x = 5\) | \(25 - 16 > 0\) | \(+\) | Increasing ↑ |
Conclusion:
Find the second derivative:
\[f''(x) = 2x\]Find inflection points by setting \(f''(x) = 0\):
\[2x = 0\] \[x = 0\]Create a sign chart for \(f''(x)\):
| Interval | Test Point | Sign of \(2x\) | Concavity |
|---|---|---|---|
| \((-\infty, 0)\) | \(x = -1\) | \(-\) | Concave Down ∩ |
| \((0, \infty)\) | \(x = 1\) | \(+\) | Concave Up ∪ |
Conclusion:
Domain:
\((-\infty, \infty)\) (all real numbers, since this is a polynomial)
x-intercepts: Solve \(f(x) = 0\)
\[\frac{x^3}{3} - 16x = 0\] \[\frac{x}{3}(x^2 - 48) = 0\] \[x = 0, \quad x = \sqrt{48} = 4\sqrt{3}, \quad x = -\sqrt{48} = -4\sqrt{3}\]y-intercept:
\[f(0) = 0\]Vertical Asymptotes:
None (polynomial functions have no vertical asymptotes)
End Behavior:
\[\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{x^3}{3} - 16x = +\infty\] \[\lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} \frac{x^3}{3} - 16x = -\infty\]Key Points:
Calculate function values at critical points:
\[f(-4) = \frac{(-4)^3}{3} - 16(-4) = \frac{-64}{3} + 64 = \frac{-64 + 192}{3} = \frac{128}{3} \approx 42.67\] \[f(4) = \frac{4^3}{3} - 16(4) = \frac{64}{3} - 64 = \frac{64 - 192}{3} = -\frac{128}{3} \approx -42.67\]Sketch Description: The graph of \(f(x) = \frac{x^3}{3} - 16x\) passes through the origin and has x-intercepts at approximately \(-6.93\), \(0\), and \(6.93\). It has a local maximum at \((-4, 128/3)\) and a local minimum at \((4, -128/3)\). The function is concave down for \(x < 0\) and concave up for \(x > 0\), with an inflection point at the origin. As \(x \to -\infty\), \(f(x) \to -\infty\), and as \(x \to \infty\), \(f(x) \to +\infty\).
Step 1: Domain
The denominator cannot be zero:
\[x^2 - 1 \neq 0 \implies x \neq \pm 1\]Domain: \((-\infty, -1) \cup (-1, 1) \cup (1, \infty)\)
Or written as: \((-\infty, \infty) \setminus \{-1, 1\}\)
Step 2: Intercepts
x-intercept: Set \(f(x) = 0\)
\[\frac{x^3}{x^2 - 1} = 0 \implies x^3 = 0 \implies x = 0\]y-intercept:
\[f(0) = \frac{0}{0 - 1} = 0\]Step 3: Vertical Asymptotes
Vertical asymptotes occur where the denominator is zero:
Vertical asymptotes at \(x = 1\) and \(x = -1\)
Behavior near asymptotes:
Step 4: End Behavior and Slant Asymptote
For rational functions, when the degree of the numerator is exactly one more than the degree of the denominator, there is a slant asymptote.
Since the degree of \(x^3\) is 3 and the degree of \(x^2 - 1\) is 2, we have \(3 = 2 + 1\), so there is a slant asymptote.
When \(x\) is large, \(\frac{x^3}{x^2 - 1} \approx \frac{x^3}{x^2} = x\)
Slant asymptote: \(y = x\)
\[\lim_{x \to \infty} f(x) = \lim_{x \to \infty} x = +\infty\] \[\lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} x = -\infty\]Step 5: First Derivative - Increasing/Decreasing
Using the quotient rule:
\[f'(x) = \frac{d}{dx}\left[\frac{x^3}{x^2 - 1}\right] = \frac{3x^2(x^2 - 1) - x^3(2x)}{(x^2 - 1)^2}\] \[= \frac{3x^4 - 3x^2 - 2x^4}{(x^2 - 1)^2} = \frac{x^4 - 3x^2}{(x^2 - 1)^2} = \frac{x^2(x^2 - 3)}{(x^2 - 1)^2}\]Critical points: Set numerator equal to zero
\[x^2(x^2 - 3) = 0\] \[x = 0, \quad x = \sqrt{3}, \quad x = -\sqrt{3}\]Sign chart for \(f'(x)\):
The numerator \(x^2(x^2 - 3)\) changes sign at \(x = \pm\sqrt{3}\). Note that \(x^2 \geq 0\) always.
The denominator \((x^2 - 1)^2 > 0\) for all \(x \neq \pm 1\).
| Interval | Sign of \((x^2 - 3)\) | Sign of \(f'(x)\) | Behavior |
|---|---|---|---|
| \((-\infty, -\sqrt{3})\) | \(+\) | \(+\) | Increasing ↑ |
| \((-\sqrt{3}, -1)\) | \(-\) | \(-\) | Decreasing ↓ |
| \((-1, 0)\) | \(-\) | \(-\) | Decreasing ↓ |
| \((0, 1)\) | \(-\) | \(-\) | Decreasing ↓ |
| \((1, \sqrt{3})\) | \(-\) | \(-\) | Decreasing ↓ |
| \((\sqrt{3}, \infty)\) | \(+\) | \(+\) | Increasing ↑ |
Conclusions:
Step 6: Second Derivative - Concavity
Finding \(f''(x)\) involves complex algebra using the quotient rule on \(f'(x)\):
\[f''(x) = \frac{x^4 - 3x^2}{(x^2 - 1)^2}\]After differentiation and simplification:
\[f''(x) = \frac{(x^2 - 1)(x)(x^2 + 6)}{(x^2 - 1)^4} = \frac{x(x^2 + 6)}{(x^2 - 1)^3}\]Setting \(f''(x) = 0\):
\[x(x^2 + 6) = 0\]Since \(x^2 + 6 > 0\) for all \(x\), we only have \(x = 0\)
Sign chart for \(f''(x)\):
| Interval | Sign of numerator \(x(x^2+6)\) | Sign of \((x^2-1)^3\) | Sign of \(f''(x)\) | Concavity |
|---|---|---|---|---|
| \((-\infty, -1)\) | \(-\) | \(+\) | \(-\) | Concave Down ∩ |
| \((-1, 0)\) | \(-\) | \(-\) | \(+\) | Concave Up ∪ |
| \((0, 1)\) | \(+\) | \(-\) | \(-\) | Concave Down ∩ |
| \((1, \infty)\) | \(+\) | \(+\) | \(+\) | Concave Up ∪ |
Conclusions:
Complete Sketch Description: The graph of \(f(x) = \frac{x^3}{x^2 - 1}\) has vertical asymptotes at \(x = -1\) and \(x = 1\), and a slant asymptote \(y = x\). The function passes through the origin, which is both an x-intercept and a y-intercept, and also an inflection point. There is a local maximum at approximately \((-\sqrt{3}, -3.9)\) and a local minimum at approximately \((\sqrt{3}, 3.9)\). The function approaches the slant asymptote \(y = x\) as \(x \to \pm\infty\).
Key Steps for Curve Sketching:
Special Note on Slant Asymptotes:
For a rational function \(f(x) = \frac{p(x)}{q(x)}\), if the degree of \(p(x)\) is exactly one more than the degree of \(q(x)\), then the function has a slant (oblique) asymptote. When \(x\) is large, the function approaches a linear function.