Session 16 - Curve Sketching

1. Name the Governing Formulas: The Polynomial Domain Theorem, Axis Intercept Formulas, and Leading Term End-Behavior Test: (These principles are formalized in Summary of Curve Sketching.)
2. Determine Algebraic Domain: Polynomials contain no division by zero, even roots, or logarithms.
   $$\text{Domain: } (-\infty, \infty)$$
3. Calculate Intercepts:
   • $y$-intercept: Set $x = 0$:
     $$f(0) = \frac{1}{4}(0)^4 - (0)^3 = 0 \implies (0, 0)$$
   • $x$-intercepts: Set $f(x) = 0$ and factor out the greatest common factor $x^3$:
     $$\frac{1}{4}x^4 - x^3 = 0 \implies x^3\left(\frac{1}{4}x - 1\right) = 0$$
     This yields:
     $$x^3 = 0 \implies x = 0$$
     $$\frac{1}{4}x - 1 = 0 \implies \frac{1}{4}x = 1 \implies x = 4$$
     The graph crosses or touches the horizontal axis at $(0, 0)$ and $(4, 0)$.
4. Test Coordinate Symmetry: Substitute $-x$ into $f(x)$:
   $$f(-x) = \frac{1}{4}(-x)^4 - (-x)^3 = \frac{1}{4}x^4 + x^3$$
   Comparing to $f(x)$:
   $$f(-x) \neq f(x) \quad \text{and} \quad f(-x) \neq -f(x)$$
   The function possesses neither even nor odd symmetry.
5. Analyze Asymptotes & End Behavior:
   • Asymptotes: None. Polynomial functions have no vertical, horizontal, or slant asymptotes.
   • End Behavior: Governed by the highest-degree term $\frac{1}{4}x^4$:
     $$\lim_{x \to \infty} \left(\frac{1}{4}x^4 - x^3\right) = \infty \quad \text{and} \quad \lim_{x \to -\infty} \left(\frac{1}{4}x^4 - x^3\right) = \infty$$
     Both the far-left and far-right tails rise without bound toward positive infinity ($\nwarrow, \nearrow$).

• (a) A vertical asymptote occurs at $x = c$ where the denominator vanishes while the numerator is non-zero, making the function value undefined. By definition, no coordinate $(c, f(c))$ can exist on the curve, so the graph cannot cross a vertical asymptote. In contrast, a horizontal or slant asymptote represents the limiting behavior of the function as $x \to \pm\infty$. The distance $|f(x) - (mx + b)|$ approaches zero as $x$ grows large, but for finite values of $x$, $f(x)$ may equal $mx + b$ wherever the remainder term $R(x) = 0$.
• (b) For any function where $f(0) = 0$, substituting $x = 0$ yields an output of $0$, establishing $(0, 0)$ as the $y$-intercept. Simultaneously, setting $f(x) = 0$ is satisfied by $x = 0$, making $(0, 0)$ a root ($x$-intercept). The origin is the unique coordinate point in the Cartesian plane that lies simultaneously on both coordinate axes.
• (c) In highway civil engineering, stationing distances are defined along the active physical construction corridor starting at entrance station $x = 0$ ($x \ge 0$). Because the singularity $x = -10$ corresponds to a station $1,000\text{ meters}$ behind the project terminus in the negative direction, the rational function profile is smooth, continuous, and well-behaved across the entire physical project domain $[0, \infty)$.

1. Name the Governing Formulas: The Rational Domain Restriction Rule, Intercept Conditions, and Slant Asymptote Long Division Formula: (Defined in Slant Asymptotes and Polynomial Division.)
2. Determine Domain: Set denominator to zero:
   $$x - 1 = 0 \implies x = 1$$
   $$\text{Domain: } (-\infty, 1) \cup (1, \infty)$$
3. Calculate Intercepts:
   • $y$-intercept: Set $x = 0$:
     $$f(0) = \frac{(0)^2 - (0) - 2}{0 - 1} = \frac{-2}{-1} = 2 \implies (0, 2)$$
   • $x$-intercepts: Set the numerator to zero and factor:
     $$x^2 - x - 2 = 0 \implies (x - 2)(x + 1) = 0 \implies x = 2 \quad \text{or} \quad x = -1$$
     Both roots belong to the domain ($x \neq 1$), yielding intercepts $(2, 0)$ and $(-1, 0)$.
4. Evaluate Vertical Asymptote: At $x = 1$, the numerator is $(1)^2 - (1) - 2 = -2 \neq 0$.
   $$\lim_{x \to 1^+} \frac{x^2 - x - 2}{x - 1} = \frac{-2}{0^+} = -\infty \quad \text{and} \quad \lim_{x \to 1^-} \frac{x^2 - x - 2}{x - 1} = \frac{-2}{0^-} = +\infty$$
   Thus, the line $x = 1$ is a vertical asymptote.
5. Extract Slant Asymptote via Polynomial Long Division: Divide $(x^2 - x - 2)$ by $(x - 1)$:
   $$\frac{x^2 - x - 2}{x - 1} = \frac{x(x - 1) - 2}{x - 1} = x - \frac{2}{x - 1}$$
   As $x \to \pm\infty$, the remainder term $-\frac{2}{x - 1} \to 0$. Therefore, the slant asymptote is the linear equation:
   $$y = x$$

1. Name the Governing Formulas: The Power Rule for Differentiation, the First Derivative Test for Local Extrema, and the Second Derivative Inflection Criterion: (These principles are formalized in First Derivative Test and Concavity and the Second Derivative Test.)
2. First Derivative & Critical Numbers: Differentiate using the Power Rule:
   $$f'(x) = \frac{d}{dx}[x^3 - 3x] = 3x^2 - 3 = 3(x^2 - 1) = 3(x - 1)(x + 1)$$
   Set $f'(x) = 0$ to find critical numbers:
   $$3(x - 1)(x + 1) = 0 \implies x = -1 \quad \text{and} \quad x = 1$$
3. First Derivative Sign Chart & Intervals of Monotonicity: Partition $(-\infty, \infty)$ using critical numbers $x = -1$ and $x = 1$:
   • Interval $(-\infty, -1)$: Test $x = -2 \implies f'(-2) = 3((-2)^2 - 1) = 9 > 0 \implies$ Increasing ($\nearrow$).
   • Interval $(-1, 1)$: Test $x = 0 \implies f'(0) = 3(0 - 1) = -3 < 0 \implies$ Decreasing ($\searrow$).
   • Interval $(1, \infty)$: Test $x = 2 \implies f'(2) = 3((2)^2 - 1) = 9 > 0 \implies$ Increasing ($\nearrow$).
4. Classify Local Extrema:
   • At $x = -1$: $f'$ changes from $+$ to $-\implies$ Local Maximum:
     $$f(-1) = (-1)^3 - 3(-1) = -1 + 3 = 2 \implies (-1, 2)$$
   • At $x = 1$: $f'$ changes from $-$ to $+\implies$ Local Minimum:
     $$f(1) = (1)^3 - 3(1) = 1 - 3 = -2 \implies (1, -2)$$
5. Second Derivative & Concavity Analysis: Differentiate $f'(x)$:
   $$f''(x) = \frac{d}{dx}[3x^2 - 3] = 6x$$
   Set $f''(x) = 0$:
   $$6x = 0 \implies x = 0$$
   Construct the concavity sign chart:
   • Interval $(-\infty, 0)$: Test $x = -1 \implies f''(-1) = 6(-1) = -6 < 0 \implies$ Concave Downward ($\cap$).
   • Interval $(0, \infty)$: Test $x = 1 \implies f''(1) = 6(1) = 6 > 0 \implies$ Concave Upward ($\cup$).
6. Identify Inflection Points: Because $f''(x)$ changes sign across $x = 0$ and $x = 0$ is in the domain:
   $$f(0) = (0)^3 - 3(0) = 0$$
   The curve has a Point of Inflection at the origin $(0, 0)$.

• (a) The Second Derivative Test states that if $f'(c) = 0$, the sign of $f''(c)$ classifies the stationary point ($>0$ for local min, $<0$ for local max). However, if $f''(c) = 0$ or does not exist, the test is completely inconclusive. For example, $f(x) = x^4$ has $f''(0) = 0$ and is a local minimum, $g(x) = -x^4$ has $g''(0) = 0$ and is a local maximum, while $h(x) = x^3$ has $h''(0) = 0$ and is an inflection point. In all such cases, you must revert to the First Derivative Test by checking the sign flip of $f'(x)$ across $c$.
• (b) An inflection point is strictly defined by geometric curvature: the curve must be continuous at $c$ and the second derivative $f''(x)$ must change sign from positive to negative or negative to positive across $x = c$. It does not require a horizontal tangent ($f'(c) = 0$). In Worked Example A, the tangent line at $(0, 0)$ has slope $f'(0) = -3$; the curve cuts through this diagonal tangent line as its curvature flips from concave down to concave up.
• (c) In exothermic reactor modeling, the second derivative represents the thermal acceleration: $\frac{d^2T}{dt^2} = \frac{d}{dt}\left[\frac{dT}{dt}\right]$. When the concavity transitions from concave up ($\frac{d^2T}{dt^2} > 0$) to concave down ($\frac{d^2T}{dt^2} < 0$) at the inflection point, the heating rate $\frac{dT}{dt}$ achieves its global mathematical peak. This instant produces the highest rate of enthalpy generation, demanding maximum coolant circulation capacity to avoid runaway vessel rupture.

1. Name the Governing Formulas: The Quotient Rule for Differentiation, the First Derivative Test, and the Second Derivative Inflection Criterion: (Defined in First Derivative Test and Concavity and the Second Derivative Test.)
2. First Derivative & Critical Numbers:
   $$g'(x) = \frac{(1)(x^2 + 1) - (x)(2x)}{(x^2 + 1)^2} = \frac{x^2 + 1 - 2x^2}{(x^2 + 1)^2} = \frac{1 - x^2}{(x^2 + 1)^2}$$
   Set $g'(x) = 0$:
   $$1 - x^2 = 0 \implies x = 1 \quad \text{or} \quad x = -1$$
   Because the denominator $(x^2 + 1)^2 \ge 1$ for all real $x$, there are no undefined critical numbers.
3. Monotonicity & Extrema Classification: The denominator is always positive, so the sign of $g'(x)$ is determined by $(1 - x^2)$:
   • On $(-\infty, -1)$: $1 - (-2)^2 = -3 < 0 \implies$ Decreasing ($\searrow$).
   • On $(-1, 1)$: $1 - (0)^2 = 1 > 0 \implies$ Increasing ($\nearrow$).
   • On $(1, \infty)$: $1 - (2)^2 = -3 < 0 \implies$ Decreasing ($\searrow$). Extrema coordinates:
   • At $x = -1$: $g'$ changes $- \to +\implies$ Local Minimum at $\left(-1, \frac{-1}{1+1}\right) = (-1, -0.5)$.
   • At $x = 1$: $g'$ changes $+ \to -\implies$ Local Maximum at $\left(1, \frac{1}{1+1}\right) = (1, 0.5)$.
4. Second Derivative & Inflection Points: Differentiate $g'(x) = \frac{1 - x^2}{(x^2 + 1)^2}$ using the Quotient Rule:
   $$g''(x) = \frac{(-2x)(x^2 + 1)^2 - (1 - x^2)\cdot 2(x^2 + 1)(2x)}{(x^2 + 1)^4}$$
   Factor out $2x(x^2 + 1)$ from the numerator:
   $$\text{Num} = 2x(x^2 + 1)\Big[ -(x^2 + 1) - 2(1 - x^2) \Big] = 2x(x^2 + 1)\Big[ -x^2 - 1 - 2 + 2x^2 \Big] = 2x(x^2 + 1)(x^2 - 3)$$
   Cancel one power of $(x^2 + 1)$:
   $$g''(x) = \frac{2x(x^2 - 3)}{(x^2 + 1)^3}$$
   Set $g''(x) = 0$:
   $$2x(x^2 - 3) = 0 \implies x = 0, \quad x = \sqrt{3}, \quad x = -\sqrt{3}$$
5. Concavity Sign Chart: The denominator $(x^2 + 1)^3 > 0$ is always positive. Testing points:
   • On $(-\infty, -\sqrt{3})$: Test $x = -2 \implies g''(-2) = \frac{2(-2)(4 - 3)}{(5)^3} = \frac{-4}{125} < 0 \implies$ Concave Down ($\cap$).
   • On $(-\sqrt{3}, 0)$: Test $x = -1 \implies g''(-1) = \frac{2(-1)(1 - 3)}{(2)^3} = \frac{4}{8} > 0 \implies$ Concave Up ($\cup$).
   • On $(0, \sqrt{3})$: Test $x = 1 \implies g''(1) = \frac{2(1)(1 - 3)}{(2)^3} = \frac{-4}{8} < 0 \implies$ Concave Down ($\cap$).
   • On $(\sqrt{3}, \infty)$: Test $x = 2 \implies g''(2) = \frac{2(2)(4 - 3)}{(5)^3} = \frac{4}{125} > 0 \implies$ Concave Up ($\cup$). Because concavity flips across all three roots, the curve has three inflection points:
     $$\left(-\sqrt{3}, -\frac{\sqrt{3}}{4}\right), \quad (0, 0), \quad \left(\sqrt{3}, \frac{\sqrt{3}}{4}\right)$$

1. Name the Governing Protocol: The Master 7-Step Curve Sketching Protocol: (Codified in Summary of Curve Sketching.)
2. Consolidate Calculated Attributes:
   • Domain: $(-\infty, \infty)$
   • Symmetry: Odd ($f(-x) = -f(x)$, origin rotation)
   • Intercepts: $(0, 0)$, $(\sqrt{3}, 0) \approx (1.73, 0)$, $(-\sqrt{3}, 0) \approx (-1.73, 0)$
   • Asymptotes: None; End behavior: $\lim_{x \to -\infty} f(x) = -\infty$, $\lim_{x \to \infty} f(x) = \infty$
   • Critical Numbers & Extrema: Local Max at $(-1, 2)$; Local Min at $(1, -2)$
   • Inflection Points: Point of inflection at $(0, 0)$
3. Construct Information Consolidation Table:

4. Layered Drafting Instructions:
   • Layer 1: No dashed asymptote lines needed.
   • Layer 2: Plot solid anchor dots at $(-1.73, 0)$, $(-1, 2)$, $(0, 0)$, $(1, -2)$, and $(1.73, 0)$.
   • Layer 3: Start in quadrant III rising up through $(-1.73, 0)$ with concave-down curvature. Peak at $(-1, 2)$ with a flat horizontal tangent. Descend steeply through $(0, 0)$, switching curvature to concave-up at the origin. Trough at $(1, -2)$ with a flat horizontal tangent, then curve upward through $(1.73, 0)$ toward $+\infty$.

• (a) The Information Consolidation Table sequences every calculated coordinate milestone (intercepts, extrema, inflection points, asymptotes) along the $x$-axis from left to right alongside its explicit first- and second-derivative instructions. This eliminates geometric sketching contradictions—such as drawing a curve that descends when the derivative is positive, drawing a peak where the second derivative indicates a valley, or inadvertently crossing a vertical boundary.
• (b) The four fundamental geometric arcs represent the four combinations of slope and curvature:
  1. Increasing & Concave Up ($\nearrow, \cup$): Tangent slopes are positive and growing steeper; the curve scoops upward with accelerating vertical climb.
  2. Increasing & Concave Down ($\nearrow, \cap$): Tangent slopes are positive but flattening toward zero; the curve arches upward toward a summit with decelerating climb.
  3. Decreasing & Concave Down ($\searrow, \cap$): Tangent slopes are negative and growing more steeply downward; the curve sheds downward with accelerating descent.
  4. Decreasing & Concave Up ($\searrow, \cup$): Tangent slopes are negative but flattening toward zero; the curve troughs or flares outward toward a horizontal floor or asymptote.
• (c) In pharmacokinetic modeling, oral medications are cleared from the circulatory system via hepatic metabolism and renal excretion. As elapsed time $t \to \infty$, the elimination rate outpaces absorption, causing the biexponential concentration profile $C(t) = 20(e^{-0.5t} - e^{-2t})$ to decay toward zero. The horizontal asymptote $C = 0$ represents complete systemic drug clearance from the patient's bloodstream.

1. Name the Governing Protocol: The Master 7-Step Curve Sketching Protocol: (Codified in Summary of Curve Sketching.)
2. Step 1: Domain & Continuity: As a polynomial, the domain is all real numbers:
   $$\text{Domain: } (-\infty, \infty)$$
3. Step 2: Intercepts:
   • $y$-intercept: $f(0) = \frac{1}{2}(0)^4 - 3(0)^2 = 0 \implies (0, 0)$.
   • $x$-intercepts: Set $f(x) = 0$:
     $$x^2\left(\frac{1}{2}x^2 - 3\right) = 0 \implies x = 0 \quad \text{or} \quad x^2 = 6 \implies x = \pm\sqrt{6} \approx \pm 2.45$$
     Intercepts: $(0, 0)$, $(\sqrt{6}, 0)$, and $(-\sqrt{6}, 0)$.
4. Step 3: Symmetry:
   $$f(-x) = \frac{1}{2}(-x)^4 - 3(-x)^2 = \frac{1}{2}x^4 - 3x^2 = f(x)$$
   The function is even; the graph is completely symmetric across the vertical $y$-axis.
5. Step 4: Asymptotes & End Behavior: No vertical, horizontal, or slant asymptotes. As $x \to \pm\infty$, the leading term $\frac{1}{2}x^4 \to \infty$. Both arms open upward toward $+\infty$.
6. Step 5: First Derivative (Monotonicity & Extrema):
   $$f'(x) = 2x^3 - 6x = 2x(x^2 - 3)$$
   Setting $f'(x) = 0$ yields critical numbers: $x = -\sqrt{3} \approx -1.73$, $x = 0$, and $x = \sqrt{3} \approx 1.73$.
   • On $(-\infty, -\sqrt{3})$: $f'(x) < 0 \implies$ Decreasing ($\searrow$).
   • On $(-\sqrt{3}, 0)$: $f'(x) > 0 \implies$ Increasing ($\nearrow$).
   • On $(0, \sqrt{3})$: $f'(x) < 0 \implies$ Decreasing ($\searrow$).
   • On $(\sqrt{3}, \infty)$: $f'(x) > 0 \implies$ Increasing ($\nearrow$). Extrema coordinates:
   • Relative Minimums: At $x = \pm\sqrt{3}$, $f(\pm\sqrt{3}) = \frac{1}{2}(9) - 3(3) = 4.5 - 9 = -4.5 \implies (-\sqrt{3}, -4.5)$ and $(\sqrt{3}, -4.5)$.
   • Relative Maximum: At $x = 0$, $f(0) = 0 \implies (0, 0)$.
7. Step 6: Second Derivative (Concavity & Inflection Points):
   $$f''(x) = 6x^2 - 6 = 6(x^2 - 1) = 6(x - 1)(x + 1)$$
   Setting $f''(x) = 0 \implies x = -1$ and $x = 1$.
   • On $(-\infty, -1) \cup (1, \infty)$: $f''(x) > 0 \implies$ Concave Upward ($\cup$).
   • On $(-1, 1)$: $f''(x) < 0 \implies$ Concave Downward ($\cap$). Inflection coordinates:
     $$f(\pm 1) = \frac{1}{2}(1) - 3(1) = -2.5 \implies (-1, -2.5) \quad \text{and} \quad (1, -2.5)$$
8. Step 7: Information Consolidation Table & Final Sketch:

Draw a smooth, symmetric "W"-shaped curve flowing through these coordinates:

1. Name the Governing Protocol: The Master 7-Step Curve Sketching Protocol: (Codified in Summary of Curve Sketching.)
2. Step 1: Domain & Continuity: Both $(3 - x)$ and $e^x$ are defined everywhere:
   $$\text{Domain: } (-\infty, \infty)$$
3. Step 2: Intercepts:
   • $y$-intercept: $g(0) = (3 - 0)e^0 = 3(1) = 3 \implies (0, 3)$.
   • $x$-intercept: $(3 - x)e^x = 0$. Since $e^x > 0$ for all real $x$, $3 - x = 0 \implies x = 3 \implies (3, 0)$.
4. Step 3: Symmetry:
   $$g(-x) = (3 - (-x))e^{-x} = (3 + x)e^{-x} \neq \pm g(x)$$
   No symmetry.
5. Step 4: Asymptotes & End Behavior:
   • No vertical asymptotes.
   • Right End Behavior ($x \to \infty$):
     $$\lim_{x \to \infty} (3 - x)e^x = (-\infty)(\infty) = -\infty$$
   • Left End Behavior ($x \to -\infty$):
     $$\lim_{x \to -\infty} (3 - x)e^x = \lim_{x \to -\infty} \frac{3 - x}{e^{-x}} \quad \left[\frac{\infty}{\infty}\right]$$
     Apply L'Hôpital's Rule:
     $$\lim_{x \to -\infty} \frac{-1}{-e^{-x}} = \lim_{x \to -\infty} \frac{1}{e^{-x}} = \lim_{x \to -\infty} e^x = 0$$
     This establishes a Horizontal Asymptote at $y = 0$ along the negative $x$-axis.
6. Step 5: First Derivative (Monotonicity & Extrema): Apply the Product Rule:
   $$g'(x) = (-1)e^x + (3 - x)e^x = (2 - x)e^x$$
   Setting $g'(x) = 0 \implies 2 - x = 0 \implies x = 2$.
   • On $(-\infty, 2)$: $g'(x) > 0 \implies$ Increasing ($\nearrow$).
   • On $(2, \infty)$: $g'(x) < 0 \implies$ Decreasing ($\searrow$).
   • Relative Maximum: Changes from increasing to decreasing at $x = 2$:
     $$g(2) = (3 - 2)e^2 = e^2 \approx 7.39 \implies (2, e^2)$$
7. Step 6: Second Derivative (Concavity & Inflection Points): Differentiate $g'(x) = (2 - x)e^x$:
   $$g''(x) = (-1)e^x + (2 - x)e^x = (1 - x)e^x$$
   Setting $g''(x) = 0 \implies 1 - x = 0 \implies x = 1$.
   • On $(-\infty, 1)$: $g''(x) > 0 \implies$ Concave Upward ($\cup$).
   • On $(1, \infty)$: $g''(x) < 0 \implies$ Concave Downward ($\cap$).
   • Point of Inflection: Concavity flips at $x = 1$:
     $$g(1) = (3 - 1)e^1 = 2e \approx 5.44 \implies (1, 2e)$$
8. Step 7: Information Consolidation Table & Final Sketch:

Explore the interactive model in Desmos:

Annotated solution curve:

1. Name the Governing Protocol: The Master 7-Step Curve Sketching Protocol: (Codified in Summary of Curve Sketching.)
2. Step 1: Domain & Discontinuities: Set denominator to zero:
   $$x^2 - 4 = 0 \implies x = \pm 2$$
   $$\text{Domain: } (-\infty, -2) \cup (-2, 2) \cup (2, \infty)$$
3. Step 2: Intercepts:
   • $y$-intercept: $h(0) = \frac{4(0)^2}{0 - 4} = 0 \implies (0, 0)$.
   • $x$-intercept: $4x^2 = 0 \implies x = 0 \implies (0, 0)$.
4. Step 3: Symmetry:
   $$h(-x) = \frac{4(-x)^2}{(-x)^2 - 4} = \frac{4x^2}{x^2 - 4} = h(x)$$
   The function is even; symmetric with respect to the $y$-axis.
5. Step 4: Asymptotes & End Behavior:
   • Vertical Asymptotes: At $x = -2$ and $x = 2$, the numerator is $4(4) = 16 \neq 0$:
     $$\text{Vertical Asymptotes at } x = -2 \quad \text{and} \quad x = 2$$
     • As $x \to 2^+$, $h(x) \to +\infty$; as $x \to 2^-$, $h(x) \to -\infty$.
     • As $x \to -2^+$, $h(x) \to -\infty$; as $x \to -2^-$, $h(x) \to +\infty$.
   • Horizontal Asymptote: Equal degrees ($n = m = 2$):
     $$\lim_{x \to \pm\infty} \frac{4x^2}{x^2 - 4} = \lim_{x \to \pm\infty} \frac{4}{1 - \frac{4}{x^2}} = 4 \implies y = 4$$
6. Step 5: First Derivative (Monotonicity & Extrema): Quotient Rule:
   $$h'(x) = \frac{(8x)(x^2 - 4) - (4x^2)(2x)}{(x^2 - 4)^2} = \frac{8x^3 - 32x - 8x^3}{(x^2 - 4)^2} = \frac{-32x}{(x^2 - 4)^2}$$
   Setting $h'(x) = 0 \implies -32x = 0 \implies x = 0$. Denominator is always positive when defined ($x \neq \pm 2$):
   • On $(-\infty, -2) \cup (-2, 0)$: $h'(x) > 0 \implies$ Increasing ($\nearrow$).
   • On $(0, 2) \cup (2, \infty)$: $h'(x) < 0 \implies$ Decreasing ($\searrow$).
   • Relative Maximum: Changes $+ \to -$ at $x = 0 \implies (0, 0)$.
7. Step 6: Second Derivative (Concavity & Inflection Points): Differentiate $h'(x) = \frac{-32x}{(x^2 - 4)^2}$:
   $$h''(x) = \frac{(-32)(x^2 - 4)^2 - (-32x)\cdot 2(x^2 - 4)(2x)}{(x^2 - 4)^4} = \frac{(x^2 - 4)[-32(x^2 - 4) + 128x^2]}{(x^2 - 4)^4} = \frac{96x^2 + 128}{(x^2 - 4)^3}$$
   The numerator $96x^2 + 128 > 0$ is strictly positive for all real $x$. Thus, $h''(x) \neq 0$ anywhere. Sign is dictated entirely by $(x^2 - 4)$:
   • On $(-\infty, -2) \cup (2, \infty)$: $x^2 - 4 > 0 \implies h''(x) > 0 \implies$ Concave Upward ($\cup$).
   • On $(-2, 2)$: $x^2 - 4 < 0 \implies h''(x) < 0 \implies$ Concave Downward ($\cap$).
   • Inflection Points: None. Although concavity flips across $x = \pm 2$, these values are vertical asymptotes that do not belong to the domain.
8. Step 7: Information Consolidation Table & Final Sketch: