Step 1: Domain

Polynomial → $(-\infty, \infty)$

Step 2: Intercepts

$y$-intercept: $f(0) = 0$ → $(0, 0)$

$x$-intercepts: $x^4 - 4x^3 = 0$ $x^3(x - 4) = 0$ $x = 0$ or $x = 4$ → $(0, 0)$ and $(4, 0)$

Step 3: Symmetry

$f(-x) = x^4 + 4x^3 \neq f(x)$ → No symmetry

Step 4: First derivative

$f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3)$

Critical points: $x = 0, 3$

Step 5: Sign chart for $f'$

Test points:

• $x = -1$: $f'(-1) = 4(1)(-4) = -16 < 0$
• $x = 1$: $f'(1) = 4(1)(-2) = -8 < 0$
• $x = 4$: $f'(4) = 4(16)(1) = 64 > 0$

        0         3
   ----  |  ----  |  ++++
      ↘       ↘  ▲  ↗

Decreasing on $(-\infty, 3)$, increasing on $(3, \infty)$

Local minimum at $x = 3$: $f(3) = 81 - 108 = -27$

Note: $x = 0$ is NOT an extremum (no sign change)

Step 6: Second derivative

$f''(x) = 12x^2 - 24x = 12x(x - 2)$

$f''(x) = 0$ when $x = 0$ or $x = 2$

Step 7: Concavity

Test points:

• $x = -1$: $f''(-1) = 12(1)(-3) = -36... wait, let me recalculate$
• $x = -1$: $f''(-1) = 12(-1)(-1-2) = 12(-1)(-3) = 36 > 0$
• $x = 1$: $f''(1) = 12(1)(1-2) = 12(1)(-1) = -12 < 0$
• $x = 3$: $f''(3) = 12(3)(3-2) = 36 > 0$

        0         2
   ++++  |  ----  |  ++++
      ∪       ∩       ∪

Inflection points at $x = 0$ and $x = 2$

$f(0) = 0$, $f(2) = 16 - 32 = -16$

Summary for sketch:

Key points:

• $(0, 0)$ - intercept and inflection point
• $(2, -16)$ - inflection point
• $(3, -27)$ - local minimum
• $(4, 0)$ - $x$-intercept

Behavior:

• Decreasing from $-\infty$ to $(3, -27)$
• Then increasing to $+\infty$
• Concave up, then down, then up again
• Passes through origin

Step 1: Domain

Denominator: $x^2 - 4 = (x-2)(x+2) = 0$ when $x = \pm 2$

Domain: $(-\infty, -2) \cup (-2, 2) \cup (2, \infty)$

Step 2: Intercepts

$y$-intercept: $g(0) = \frac{0}{-4} = 0$ → $(0, 0)$

$x$-intercepts: $x^2 = 0$ → $x = 0$ → $(0, 0)$

Step 3: Symmetry

$g(-x) = \frac{(-x)^2}{(-x)^2 - 4} = \frac{x^2}{x^2-4} = g(x)$

EVEN function - symmetric about $y$-axis!

Step 4: Asymptotes

Vertical: At $x = -2$ and $x = 2$ (where denominator = 0)

Horizontal: $\lim_{x \to \infty} \frac{x^2}{x^2 - 4} = \lim_{x \to \infty} \frac{1}{1 - 4/x^2} = 1$

Horizontal asymptote: $y = 1$

Step 5: First derivative (Quotient Rule)

$g'(x) = \frac{2x(x^2-4) - x^2(2x)}{(x^2-4)^2}$

$= \frac{2x^3 - 8x - 2x^3}{(x^2-4)^2}$

$= \frac{-8x}{(x^2-4)^2}$

Critical point: $-8x = 0$ → $x = 0$

Step 6: Sign of $g'$

Since $(x^2-4)^2 > 0$ always, sign depends on $-8x$:

• $x < 0$: $-8x > 0$ → increasing
• $x > 0$: $-8x < 0$ → decreasing

Local maximum at $x = 0$: $g(0) = 0$

Step 7: Second derivative (skip for brevity)

Can verify concavity, but we have enough information.

Step 8: Behavior near asymptotes

Near $x = 2^+$: numerator → 4, denominator → $0^+$ → $g(x) \to +\infty$

Near $x = 2^-$: numerator → 4, denominator → $0^-$ → $g(x) \to -\infty$

By symmetry, similar behavior at $x = -2$

Summary for sketch:

• Symmetric about $y$-axis
• Vertical asymptotes at $x = \pm 2$
• Horizontal asymptote at $y = 1$
• Local max at $(0, 0)$
• Increasing on $(-\infty, -2)$ and $(-2, 0)$
• Decreasing on $(0, 2)$ and $(2, \infty)$
• As $x \to \pm\infty$, $g(x) \to 1$

Step 1: Domain

All real numbers: $(-\infty, \infty)$

Step 2: Intercepts

$y$-intercept: $h(0) = 0 \cdot e^0 = 0$ → $(0, 0)$

$x$-intercepts: $xe^{-x} = 0$ → $x = 0$ → $(0, 0)$

Step 3: Limits (behavior at infinity)

As $x \to -\infty$: $e^{-x} \to \infty$, so $h(x) \to -\infty$

As $x \to +\infty$: $\frac{x}{e^x} \to 0$ (exponential dominates)

Horizontal asymptote: $y = 0$ (as $x \to \infty$)

Step 4: First derivative (Product Rule)

$h'(x) = (1)(e^{-x}) + (x)(-e^{-x})$

$= e^{-x} - xe^{-x}$

$= e^{-x}(1 - x)$

Critical point: $1 - x = 0$ → $x = 1$

Step 5: Sign of $h'$

$e^{-x} > 0$ always, so sign depends on $(1-x)$:

• $x < 1$: $(1-x) > 0$ → increasing
• $x > 1$: $(1-x) < 0$ → decreasing

Local maximum at $x = 1$: $h(1) = 1 \cdot e^{-1} = \frac{1}{e} \approx 0.368$

Step 6: Second derivative

$h''(x) = (-e^{-x})(1-x) + e^{-x}(-1)$

$= -e^{-x}(1-x) - e^{-x}$

$= -e^{-x}(1-x+1)$

$= -e^{-x}(2-x)$

$= e^{-x}(x-2)$

Inflection point: $x - 2 = 0$ → $x = 2$

Step 7: Concavity

• $x < 2$: $(x-2) < 0$ → concave down
• $x > 2$: $(x-2) > 0$ → concave up

Inflection point at $x = 2$: $h(2) = 2e^{-2} = \frac{2}{e^2} \approx 0.271$

Summary for sketch:

Key points:

• $(0, 0)$ - origin
• $(1, 1/e)$ - local maximum
• $(2, 2/e^2)$ - inflection point

Behavior:

• Increases from $-\infty$ to max at $x=1$
• Decreases from max, approaching 0 as $x \to \infty$
• Concave down until $x=2$, then concave up
• Horizontal asymptote $y=0$ on the right

Answer: The graph rises from negative infinity through the origin, reaches a maximum at $(1, 1/e)$, has an inflection point at $(2, 2/e^2)$, and approaches 0 as $x \to \infty$.

Step 1: Find intercepts: y-intercept: y(0) = 0 x-intercepts: x³ - 3x² = 0 → x²(x - 3) = 0 → x = 0, 3

Step 2: Find f'(x) and critical points: f'(x) = 3x² - 6x = 3x(x - 2) Critical points: x = 0, 2

Step 3: First derivative test: f'(-1) = 3(-1)(−3) = 9 > 0 (increasing) f'(1) = 3(1)(−1) = -3 < 0 (decreasing) f'(3) = 3(3)(1) = 9 > 0 (increasing) Local max at x = 0: f(0) = 0 Local min at x = 2: f(2) = 8 - 12 = -4

Step 4: Find f''(x) and inflection points: f''(x) = 6x - 6 = 6(x - 1) Inflection point: x = 1, f(1) = 1 - 3 = -2

Step 5: Concavity: f''(0) = -6 < 0 (concave down) f''(2) = 6 > 0 (concave up)

Step 6: End behavior: As x → ∞: y → ∞ (positive cubic) As x → -∞: y → -∞

Answer: Curve crosses (0,0) and (3,0), local max at (0,0), local min at (2,-4), inflection at (1,-2)

Step 1: Domain and intercepts: Domain: all real numbers (denominator never zero) x-intercept: x = 0 y-intercept: f(0) = 0

Step 2: Symmetry: f(-x) = -x/(x² + 1) = -f(x) → odd function (symmetric about origin)

Step 3: Horizontal asymptotes: lim(x→±∞) x/(x² + 1) = lim(x→±∞) (1/x)/(1 + 1/x²) = 0 Horizontal asymptote: y = 0

Step 4: Find f'(x) using quotient rule: f'(x) = [(1)(x² + 1) - x(2x)]/(x² + 1)² = (x² + 1 - 2x²)/(x² + 1)² = (1 - x²)/(x² + 1)²

Step 5: Critical points: f'(x) = 0 when 1 - x² = 0 → x = ±1 f(1) = 1/2, f(-1) = -1/2 Local max at (1, 1/2), local min at (-1, -1/2)

Step 6: Find f''(x) (or just note concavity): By symmetry and shape, inflection points exist

Answer: Odd function, passes through origin, local max (1, 1/2), local min (-1, -1/2), horizontal asymptote y = 0