Curve Sketching: The Full Checklist

$f(x)=\dfrac{(x-2)(x+2)}{(x-3)(x+3)}$ — nothing cancels, so there are no holes and $x=\pm3$ are vertical asymptotes. $f(-x)=f(x)$, so $f$ is even.

At $x=3$ the numerator is $5>0$ while $x^{2}-9$ passes from $0^{-}$ to $0^{+}$:

$\lim_{x\to3^{-}}f(x)=-\infty,\qquad \lim_{x\to3^{+}}f(x)=+\infty$

Evenness mirrors this at $x=-3$: $+\infty$ from the left, $-\infty$ from the right. Equal degrees give $y=1$. Since $f(x)-1=\dfrac{5}{x^{2}-9}$ is never $0$, the curve never crosses $y=1$: above it for $|x|>3$, below it for $|x|<3$.

Long division gives $2x^{2}+3x-1=(x+2)(2x-1)+1$, so

$f(x)=2x-1+\frac{1}{x+2}$

The numerator at $x=-2$ is $1\neq0$, so $x=-2$ is a vertical asymptote: $\frac{1}{x+2}\to+\infty$ from the right and $-\infty$ from the left, while $2x-1\to-5$ stays bounded.

Since $\frac{1}{x+2}\to0$ as $x\to\pm\infty$, $\lim_{x\to\pm\infty}[f(x)-(2x-1)]=0$: the slant asymptote is $y=2x-1$ and there is no horizontal one. The difference $\frac{1}{x+2}$ is positive for $x>-2$ (above) and negative for $x<-2$ (below).

$f'(x)=3x^{2}-6x-9=3(x+1)(x-3)$

The sign of $f'$ is $+,-,+$ across $x=-1$ and $x=3$. So $f'$ turns $+\to-$ at $x=-1$ — a local maximum, $f(-1)=-1-3+9+5=10$ — and $-\to+$ at $x=3$, a local minimum, $f(3)=27-27-27+5=-22$.

$f''(x)=6x-6=6(x-1)$ changes sign at $x=1$, and $f(1)=-6$, so $(1,-6)$ is an inflection point: concave down to its left, up to its right. Cross-check: $f''(-1)=-12<0$ and $f''(3)=12>0$ match the classification.

As $x\to\infty$, $\frac{x^{2}}{e^{x}}$ is $\frac{\infty}{\infty}$, so l'Hopital twice gives $\lim\frac{2x}{e^{x}}=\lim\frac{2}{e^{x}}=0$, approached from above since $f\ge0$. As $x\to-\infty$ both $x^{2}$ and $e^{-x}$ grow, so $f\to+\infty$ — not indeterminate, so l'Hopital is illegal there.

$f'(x)=xe^{-x}(2-x),\qquad f''(x)=e^{-x}\left(x^{2}-4x+2\right)$

As $e^{-x}>0$, $f'$ has the sign of $x(2-x)$: down, up, down. Local min $(0,0)$, local max $\left(2,4e^{-2}\right)\approx(2,0.54)$. $f''=0$ at $x=2\pm\sqrt2\approx0.59,\,3.41$, both genuine sign changes, so both give inflection points.

$y'=3x^{2}+2ax+b$ and $y''=6x+2a$.

Inflection at $x=1$ forces $y''(1)=6+2a=0$, so $a=-3$. Then $y'(2)=12+4(-3)+b=0$ gives $b=0$, and $y=x^{3}-3x^{2}+1$.

Verify: $y''=6(x-1)$ really does change sign at $x=1$, and $y(1)=-1$, so $(1,-1)$ is an inflection point.

$y'=3x(x-2)$ gives critical numbers $0$ and $2$. $y''(0)=-6<0$: local maximum $(0,1)$. $y''(2)=6>0$: local minimum $(2,-3)$.

$f(x)=x^{5/3}-5x^{2/3}$, so

$f'(x)=\frac53x^{2/3}-\frac{10}{3}x^{-1/3}=\frac{5(x-2)}{3x^{1/3}}$

$f'=0$ at $x=2$; $f'$ is undefined at $x=0$, which is in the domain, so both are critical numbers. The signs of $f'$ are $+,-,+$ on $(-\infty,0)$, $(0,2)$, $(2,\infty)$: local maximum $(0,0)$, local minimum $\left(2,-3\cdot2^{2/3}\right)$.

At $0$, $f'\to+\infty$ from the left and $-\infty$ from the right — opposite signs, so a cusp.

$f''(x)=\dfrac{10(x+1)}{9x^{4/3}}$ has the sign of $x+1$, so concavity changes only at $x=-1$: inflection point $(-1,-6)$.