Curve Sketching

A polynomial: all real $x$.

$f'(x) = 3x^2 - 12x + 9 = 3(x - 1)(x - 3)$.

$f' = 0$ at $x = 1$ and $x = 3$; $f'$ is never undefined.

$(-\infty, 1)$: $+$. $(1, 3)$: $-$. $(3, \infty)$: $+$.

Increasing on $(-\infty, 1)$ and $(3, \infty)$, decreasing on $(1, 3)$. Local max $f(1) = 5$, local min $f(3) = 1$.

$x^{2/3} = (\sqrt[3]{x})^2$ is defined for all $x$: all real $x$.

$f(x) = x^{5/3} - 5x^{2/3}$, so $f'(x) = \tfrac{5}{3}x^{2/3} - \tfrac{10}{3}x^{-1/3} = \dfrac{5(x - 2)}{3x^{1/3}}$.

$f' = 0$ at $x = 2$; $f'$ undefined at $x = 0$, which is in the domain. Critical points $0$ and $2$.

$(-\infty, 0)$: $\dfrac{-}{-} = +$. $(0, 2)$: $\dfrac{-}{+} = -$. $(2, \infty)$: $+$.

Increasing on $(-\infty, 0)$ and $(2, \infty)$, decreasing on $(0, 2)$. Local max $f(0) = 0$; local min $f(2) = -3\sqrt[3]{4} \approx -4.76$.

$x \neq 1$.

$f'(x) = \dfrac{2x(x - 1) - x^2}{(x - 1)^2} = \dfrac{x(x - 2)}{(x - 1)^2}$.

$f' = 0$ at $x = 0$ and $x = 2$. $f'$ is undefined at $x = 1$, but $1$ is not in the domain: not a critical point.

Cut at $0, 1, 2$. $(-\infty, 0)$: $+$. $(0, 1)$: $-$. $(1, 2)$: $-$. $(2, \infty)$: $+$.

Increasing on $(-\infty, 0)$ and $(2, \infty)$; decreasing on $(0, 1)$ and $(1, 2)$. Local max $f(0) = 0$, local min $f(2) = 4$.

$f'(x) = 4x^3 - 12x$, $f''(x) = 12x^2 - 12 = 12(x - 1)(x + 1)$.

$f'' = 0$ at $x = \pm 1$; never undefined.

$(-\infty, -1)$: $+$. $(-1, 1)$: $-$. $(1, \infty)$: $+$.

Concave up on $(-\infty, -1)$ and $(1, \infty)$, concave down on $(-1, 1)$.

The sign changes at both, and $f(\pm 1) = 1 - 6 + 3 = -2$: inflection points $(-1, -2)$ and $(1, -2)$.

$f'(x) = 5x^4 - 20x^3$, $f''(x) = 20x^3 - 60x^2 = 20x^2(x - 3)$.

$f'' = 0$ at $x = 0$ (a double root) and $x = 3$.

$(-\infty, 0)$: $(+)(-) = -$. $(0, 3)$: $(+)(-) = -$. $(3, \infty)$: $(+)(+) = +$.

Concave down on $(-\infty, 0)$ and $(0, 3)$, concave up on $(3, \infty)$.

No sign change at $0$, so no inflection there. At $3$ it changes: $f(3) = 243 - 405 = -162$.

$f'(x) = \tfrac{1}{3}x^{-2/3}$, $f''(x) = -\tfrac{2}{9}x^{-5/3} = -\dfrac{2}{9x^{5/3}}$.

$f''$ is never $0$; it is undefined at $x = 0$, where $f(0) = 0$ is defined.

$(-\infty, 0)$: $x^{5/3} < 0$, so $f'' > 0$. $(0, \infty)$: $f'' < 0$.

Concave up on $(-\infty, 0)$, concave down on $(0, \infty)$.

The sign changes at $0$ and $f(0) = 0$ exists: inflection point $(0, 0)$.

$f'(x) = \dfrac{(x - 2) - (x + 1)}{(x - 2)^2} = \dfrac{-3}{(x - 2)^2}$, so $f''(x) = \dfrac{6}{(x - 2)^3}$.

$f''$ is never $0$; undefined at $x = 2$, which is not in the domain.

$(-\infty, 2)$: $-$. $(2, \infty)$: $+$.

Concave down on $(-\infty, 2)$, concave up on $(2, \infty)$.

The sign changes only at $x = 2$, where $f$ is undefined (a vertical asymptote): no inflection point.

It is $f'$. Zeros at $x = 0$ (crosses) and $x = 3$ (touches); peak at $x = 1$, valley at $x = 3$.

$f' < 0$ on $(-\infty, 0)$ and $f' > 0$ on $(0, 3)$ and $(3, \infty)$: $f$ decreases on $(-\infty, 0)$ and increases on $(0, \infty)$.

At $x = 0$, $f'$ goes $-$ to $+$: local min. At $x = 3$, $f'$ touches $0$ but stays $+$: no extremum.

$f'$ rises on $(-\infty, 1)$ and $(3, \infty)$: concave up there. $f'$ falls on $(1, 3)$: concave down.

$f'$ peaks at $1$ and bottoms out at $3$: inflection points at $x = 1$ and $x = 3$.

It is $f'$. Zeros at $x = -1$ and $x = 3$ (both crossings); peak at $x = 1$.

$f' < 0$ on $(-\infty, -1)$ and $(3, \infty)$, $f' > 0$ on $(-1, 3)$: $f$ decreases, increases, decreases.

At $x = -1$: $-$ to $+$, local min. At $x = 3$: $+$ to $-$, local max.

$f'$ rises on $(-\infty, 1)$: concave up. $f'$ falls on $(1, \infty)$: concave down.

$f'$ peaks at $x = 1$: one inflection point, at $x = 1$.

Domain $x \neq \pm 2$. Only intercept $(0, 0)$. $f(-x) = f(x)$: even, symmetric about the $y$-axis.

Vertical: $x = \pm 2$, with $f \to +\infty$ as $x \to 2^+$ and $x \to -2^-$, $f \to -\infty$ as $x \to 2^-$ and $x \to -2^+$. Horizontal: $y = 1$ at both ends.

$f' = 0$ only at $x = 0$. $f' > 0$ for $x < 0$, $< 0$ for $x > 0$: increasing on $(-\infty, -2)$, $(-2, 0)$, decreasing on $(0, 2)$, $(2, \infty)$. Local max $(0, 0)$.

$3x^2 + 4 > 0$, so $f''$ has the sign of $x^2 - 4$: concave up on $(-\infty, -2)$, $(2, \infty)$; concave down on $(-2, 2)$. No inflection points (changes only at the asymptotes).

Middle: an upside-down cup through its top $(0, 0)$, dropping to $-\infty$ at $x = \pm 2$. Outside: two branches above $y = 1$, rising to $+\infty$ at the asymptotes and flattening to $y = 1$.

Domain all real $x$. $f(x) = 0$ only at $x = 0$: intercept $(0, 0)$. $f(-x) = x^2e^{x}$: no symmetry.

No vertical asymptotes. $\displaystyle\lim_{x\to\infty}\dfrac{x^2}{e^x} = 0$ (L'Hôpital twice): $y = 0$ on the right. As $x \to -\infty$, $f \to \infty$: no asymptote on the left.

$f'(x) = (2x - x^2)e^{-x} = x(2 - x)e^{-x}$. Signs: $-$, $+$, $-$ around $0$ and $2$. Local min $(0, 0)$, local max $(2, 4e^{-2}) \approx (2, 0.54)$.

$f''(x) = (x^2 - 4x + 2)e^{-x}$, zero at $x = 2 \pm \sqrt{2}$. Concave up on $(-\infty, 2 - \sqrt{2})$ and $(2 + \sqrt{2}, \infty)$, down between. Inflection points $\approx (0.59, 0.19)$ and $(3.41, 0.38)$.

From $+\infty$ on the left, fall to the min $(0, 0)$, rise (bending over at $x \approx 0.59$) to the max $(2, 0.54)$, then fall and flatten toward $y = 0$ from above, bending up at $x \approx 3.41$.