(a) No VA; horizontal asymptote A1·A1
Denominator $x^{2} + 1 \ge 1 > 0$ for every real $x$, so it never vanishes.
No vertical asymptote.
Degree of numerator ($1$) $<$ degree of denominator ($2$), so the HA is $y = 0$:
$$ \lim_{x \to \pm\infty} \frac{x}{x^{2} + 1} \;=\; \lim_{x \to \pm\infty} \frac{1/x}{1 + 1/x^{2}} \;=\; \frac{0}{1} \;=\; 0. $$
(b) Derivative by quotient rule M1·A1·A1
Let $u = x$ and $v = x^{2} + 1$. Then $u' = 1$ and $v' = 2x$. Quotient rule:
$$ f'(x) \;=\; \frac{u'v - uv'}{v^{2}} \;=\; \frac{1 \cdot (x^{2} + 1) - x \cdot 2x}{(x^{2} + 1)^{2}} \;=\; \frac{x^{2} + 1 - 2x^{2}}{(x^{2} + 1)^{2}} \;=\; \frac{1 - x^{2}}{(x^{2} + 1)^{2}}. \quad \text{AG} $$
(c) Turning points and classification M1·A1·A1
The denominator $(x^{2} + 1)^{2} > 0$ is always positive, so the sign of $f'(x)$ equals the sign of the numerator $1 - x^{2} = (1 - x)(1 + x)$.
$f'(x) = 0 \Leftrightarrow 1 - x^{2} = 0 \Leftrightarrow x = \pm 1$.
Sign of $f'$:
- $x < -1$: $(1 - x) > 0$, $(1 + x) < 0 \Rightarrow f'(x) < 0$.
- $-1 < x < 1$: $(1 - x) > 0$, $(1 + x) > 0 \Rightarrow f'(x) > 0$.
- $x > 1$: $(1 - x) < 0$, $(1 + x) > 0 \Rightarrow f'(x) < 0$.
At $x = -1$, $f'$ goes from $-$ to $+$:
local minimum. At $x = 1$, $f'$ goes from $+$ to $-$:
local maximum.
Values: $f(1) = \dfrac{1}{1 + 1} = \dfrac{1}{2}$ and $f(-1) = \dfrac{-1}{1 + 1} = -\dfrac{1}{2}$.
Turning points: $\bigl(1, \tfrac{1}{2}\bigr)$ (max), $\bigl(-1, -\tfrac{1}{2}\bigr)$ (min).
(d) Range A1
$f$ is continuous on $\mathbb{R}$, has $\lim_{x \to \pm\infty} f(x) = 0$, and attains a global maximum $\tfrac{1}{2}$ at $x = 1$ and a global minimum $-\tfrac{1}{2}$ at $x = -1$. By the intermediate-value theorem $f$ takes every value between $-\tfrac{1}{2}$ and $\tfrac{1}{2}$, and both endpoints are attained.
Range: $\bigl[-\tfrac{1}{2},\, \tfrac{1}{2}\bigr]$.
The denominator $x^{2} + 1$ is the canonical "never zero" trick. Any time you see $x^{2} + 1$ (or $x^{2} + c$ with $c > 0$) in a denominator, the function has no vertical asymptotes — and the quotient-rule derivative will inherit this clean denominator, raised to a power. That gives you a free win: the sign of $f'$ is controlled entirely by its numerator, so you skip the "test where denominator changes sign" step. Pattern-match this on Paper 2 / Paper 3: $\dfrac{p(x)}{x^{2} + 1}$, $\dfrac{p(x)}{(x^{2} + 1)^{2}}$, etc., reduce extrema questions to root-finding on the numerator. Bonus: both endpoints of the range are attained here (closed brackets), unlike Q6 where the HA contributed an open bracket. Always check whether the asymptote value gets shadowed by an interior extremum that captures the same value attained — here $y = 0$ is sandwiched between $\pm 1/2$, so it sits inside the range, not at its boundary.
(a) 无竖直渐近线；水平渐近线 A1·A1
分母 $x^{2} + 1 \ge 1 > 0$，恒不为零。
无竖直渐近线。
分子次数 $1 <$ 分母次数 $2$，故水平渐近线 $y = 0$：
$$ \lim_{x \to \pm\infty} \frac{x}{x^{2} + 1} \;=\; \lim_{x \to \pm\infty} \frac{1/x}{1 + 1/x^{2}} \;=\; \frac{0}{1} \;=\; 0. $$
(b) 商法则求导 M1·A1·A1
设 $u = x$、$v = x^{2} + 1$。则 $u' = 1$、$v' = 2x$。商法则：
$$ f'(x) \;=\; \frac{u'v - uv'}{v^{2}} \;=\; \frac{1 \cdot (x^{2} + 1) - x \cdot 2x}{(x^{2} + 1)^{2}} \;=\; \frac{x^{2} + 1 - 2x^{2}}{(x^{2} + 1)^{2}} \;=\; \frac{1 - x^{2}}{(x^{2} + 1)^{2}}. \quad \text{AG} $$
(c) 转折点与判定 M1·A1·A1
分母 $(x^{2} + 1)^{2} > 0$ 恒正，故 $f'(x)$ 的符号由分子 $1 - x^{2} = (1 - x)(1 + x)$ 决定。
$f'(x) = 0 \Leftrightarrow 1 - x^{2} = 0 \Leftrightarrow x = \pm 1$。
$f'$ 的符号：
- $x < -1$：$(1 - x) > 0$、$(1 + x) < 0 \Rightarrow f'(x) < 0$。
- $-1 < x < 1$：两因子同正 $\Rightarrow f'(x) > 0$。
- $x > 1$：$(1 - x) < 0$、$(1 + x) > 0 \Rightarrow f'(x) < 0$。
$x = -1$ 处 $f'$ 由 $-$ 变 $+$：
极小；$x = 1$ 处 $f'$ 由 $+$ 变 $-$：
极大。
值：$f(1) = \dfrac{1}{1 + 1} = \dfrac{1}{2}$，$f(-1) = -\dfrac{1}{2}$。
转折点：$\bigl(1, \tfrac{1}{2}\bigr)$（极大）、$\bigl(-1, -\tfrac{1}{2}\bigr)$（极小）。
(d) 值域 A1
$f$ 在 $\mathbb{R}$ 上连续，$\lim_{x \to \pm\infty} f(x) = 0$，全局极大 $\tfrac{1}{2}$（$x = 1$）、全局极小 $-\tfrac{1}{2}$（$x = -1$）。由介值定理 $f$ 取遍 $\bigl[-\tfrac{1}{2}, \tfrac{1}{2}\bigr]$ 中每个值，且两端点都取到。
值域：$\bigl[-\tfrac{1}{2},\, \tfrac{1}{2}\bigr]$。
分母 $x^{2} + 1$ 是经典的"恒不为零"套路。分母里出现 $x^{2} + 1$（或 $x^{2} + c$，$c > 0$）时，函数无竖直渐近线——商法则求导后分母还会更高次但仍非零。这送你一份福利：$f'$ 的符号完全由分子决定，"分母变号"那一步可以跳过。Paper 2 / Paper 3 模式匹配：$\dfrac{p(x)}{x^{2} + 1}$、$\dfrac{p(x)}{(x^{2} + 1)^{2}}$ 等，求极值就只需对分子求根。附赠：本题值域两端都取到（闭括号），与 Q6 中水平渐近线提供开括号不同。要核查：水平渐近线之值是否被某个内部极值"覆盖"——本题 $y = 0$ 夹在 $\pm 1/2$ 之间，落在值域内部，并非端点。