Corrected exercises: curve sketching (MATH 140)

The denominator $x^2 - 4 = (x - 2)(x + 2)$ vanishes at $\pm 2$, so the domain is $\mathbb{R} \setminus \{-2, 2\}$. The numerator $x^2 + 4 \ge 4$ never vanishes: no $x$-intercept. $f(0) = \frac{4}{-4} = -1$, the $y$-intercept is $(0, -1)$. Only even powers of $x$ appear, so $f(-x) = f(x)$: $f$ is even and its graph is symmetric about the $y$-axis. Symmetry halves the computations, but the table is still written on the whole domain, and the two excluded points will cut EVERY row of it. That is why the domain comes first.

At $x = 2$ the numerator tends to $8 \ne 0$ and the denominator to $0$, so the limit is infinite and its sign is the sign of $x^2 - 4$. As $x \to 2^+$, $x^2 - 4 \to 0^+$ and $f \to +\infty$; as $x \to 2^-$, $x^2 - 4 \to 0^-$ and $f \to -\infty$. By symmetry, $f \to +\infty$ as $x \to -2^-$ and $f \to -\infty$ as $x \to -2^+$. No factor cancels, so $x = 2$ and $x = -2$ are genuine vertical asymptotes. At $\pm\infty$, divide by $x^2$: $f(x) = \frac{1 + 4/x^2}{1 - 4/x^2} \to 1$, so $y = 1$ is the horizontal asymptote at both ends. Its position: $f(x) - 1 = \frac{8}{x^2 - 4} > 0$ for $|x| > 2$, so the two outer branches stay ABOVE $y = 1$ and never cross it.

Quotient rule: $f'(x) = \frac{2x(x^2 - 4) - (x^2 + 4)(2x)}{(x^2 - 4)^2} = \frac{2x\left[(x^2 - 4) - (x^2 + 4)\right]}{(x^2 - 4)^2} = \frac{-16x}{(x^2 - 4)^2}$. The denominator is positive on the domain, so $f'$ has the sign of $-16x$: positive for $x < 0$, negative for $x > 0$. The only critical number is $0$; the points $\pm 2$ are NOT critical numbers, since they are not in the domain. So $f$ increases on $(-\infty, -2)$ and on $(-2, 0]$, decreases on $[0, 2)$ and on $(2, \infty)$, and the first derivative test gives a local maximum $f(0) = -1$. Writing increasing on $(-\infty, 0]$ in one piece is false: $-3 < -1$ and yet $f(-3) = \frac{13}{5} > f(-1) = -\frac{5}{3}$. The asymptote breaks the interval, and so must the answer.

Write $f'(x) = -16x(x^2 - 4)^{-2}$ and use the product rule, then the chain rule with inner function $x^2 - 4$: $f''(x) = -16(x^2 - 4)^{-2} + (-16x)(-2)(x^2 - 4)^{-3}(2x) = \frac{-16(x^2 - 4) + 64x^2}{(x^2 - 4)^3} = \frac{48x^2 + 64}{(x^2 - 4)^3} = \frac{16(3x^2 + 4)}{(x^2 - 4)^3}$. The numerator is always positive, so $f''$ has the sign of $x^2 - 4$: concave up for $|x| > 2$, concave down for $|x| < 2$. The concavity does change at $x = \pm 2$, but an inflection point is a POINT $(a, f(a))$ of the graph where the concavity changes, and $f(2)$ does not exist. So $f$ has NO inflection point. Listing $(2, f(2))$ as an inflection point costs the mark, and the sketch cannot even place it.

The table, from left to right: on $(-\infty, -2)$, $f' > 0$ and $f'' > 0$, the branch leaves $y = 1$ from above and climbs, concave up, to $+\infty$ at $-2^-$; on $(-2, 2)$, the branch comes up from $-\infty$, reaches its maximum $(0, -1)$ and falls back to $-\infty$, concave down all the way; on $(2, \infty)$ the branch comes down from $+\infty$ toward $y = 1$, concave up. The sketch in the figure is nothing more than this table drawn. Range: the middle branch takes every value of $(-\infty, -1]$, by continuity and the values of the limits; the outer branches take every value of $(1, \infty)$, never $1$ itself. Values in $(-1, 1]$ are never taken, so the range is $(-\infty, -1] \cup (1, \infty)$. Consistency check: a middle branch drawn crossing the $x$-axis would contradict a), and outer branches dipping below $y = 1$ would contradict b).

$x^2 + 1 \ge 1 > 0$, so the domain is $\mathbb{R}$ and $f$ is continuous everywhere: no vertical asymptote, since that would need an infinite limit at a finite point. $f(0) = 3$ and $f(x) = 0 \iff x = -3$. $f(1) = \frac{4}{\sqrt 2} = 2\sqrt 2$ while $f(-1) = \frac{2}{\sqrt 2} = \sqrt 2$, neither equal nor opposite: no symmetry. At infinity, factor $x^2$ under the root: $\sqrt{x^2 + 1} = |x|\sqrt{1 + 1/x^2}$. For $x > 0$, $|x| = x$ and $f(x) = \frac{1 + 3/x}{\sqrt{1 + 1/x^2}} \to 1$. For $x < 0$, $|x| = -x$ and $f(x) = -\frac{1 + 3/x}{\sqrt{1 + 1/x^2}} \to -1$. Two different horizontal asymptotes: $y = 1$ on the right, $y = -1$ on the left. Writing $\sqrt{x^2} = x$ on both sides gives $y = 1$ twice, and the whole left half of the sketch is then wrong.

Product rule on $(x + 3)(x^2 + 1)^{-1/2}$, with the chain rule on the second factor (inner function $x^2 + 1$): $f'(x) = (x^2 + 1)^{-1/2} + (x + 3)\left(-\frac{1}{2}\right)(x^2 + 1)^{-3/2}(2x) = \frac{(x^2 + 1) - x(x + 3)}{(x^2 + 1)^{3/2}} = \frac{1 - 3x}{(x^2 + 1)^{3/2}}$. The denominator is positive, so $f'$ has the sign of $1 - 3x$. The only critical number is $\frac{1}{3}$, where $f'$ goes from $+$ to $-$. Since $f$ increases on all of $\left(-\infty, \frac{1}{3}\right]$ and decreases on all of $\left[\frac{1}{3}, \infty\right)$, this local maximum is the absolute maximum: $f\left(\frac{1}{3}\right) = \frac{10/3}{\sqrt{10/9}} = \frac{10}{3} \cdot \frac{3}{\sqrt{10}} = \sqrt{10}$. There is no minimum: the values come arbitrarily close to $-1$ on the left without reaching it.

$f(x) = 1 \iff x + 3 = \sqrt{x^2 + 1}$. The right side is positive, so a solution needs $x + 3 \ge 0$. Squaring: $x^2 + 6x + 9 = x^2 + 1$, so $6x = -8$ and $x = -\frac{4}{3}$, which satisfies $x + 3 = \frac{5}{3} \ge 0$ and is kept (check: $\sqrt{\frac{16}{9} + 1} = \frac{5}{3}$). The graph crosses the line $y = 1$ exactly once, at $x = -\frac{4}{3}$, on the way UP, far to the left of the region where $y = 1$ is an asymptote. There is no contradiction: a horizontal asymptote describes the behaviour as $x \to \infty$ only, and says nothing about finite $x$. After the crossing $f > 1$ for good, so the curve climbs to $\sqrt{10}$ and then comes back down toward $y = 1$ from above, never touching it again.

From $f'(x) = (1 - 3x)(x^2 + 1)^{-3/2}$, product rule and chain rule: $f''(x) = -3(x^2 + 1)^{-3/2} + (1 - 3x)\left(-\frac{3}{2}\right)(2x)(x^2 + 1)^{-5/2} = \frac{-3(x^2 + 1) - 3x(1 - 3x)}{(x^2 + 1)^{5/2}} = \frac{6x^2 - 3x - 3}{(x^2 + 1)^{5/2}} = \frac{3(2x + 1)(x - 1)}{(x^2 + 1)^{5/2}}$. Signs: $+$ for $x < -\frac{1}{2}$, $-$ on $\left(-\frac{1}{2}, 1\right)$, $+$ for $x > 1$. Both points are in the domain and $f''$ changes sign there, so both are inflection points: $f\left(-\frac{1}{2}\right) = \frac{5/2}{\sqrt{5/4}} = \frac{5}{2} \cdot \frac{2}{\sqrt 5} = \sqrt 5$ and $f(1) = \frac{4}{\sqrt 2} = 2\sqrt 2$.

Reading the table from left to right: the curve leaves $y = -1$ from above, rising and concave up; crosses $y = 0$ at $-3$ and $y = 1$ at $-\frac{4}{3}$; turns concave down at $\left(-\frac{1}{2}, \sqrt 5\right)$; reaches its top $\left(\frac{1}{3}, \sqrt{10}\right)$; turns concave up again at $(1, 2\sqrt 2)$ while decreasing; and settles on $y = 1$ from above. The values agree with each other: $\sqrt 5 < \sqrt{10}$ and $2\sqrt 2 = \sqrt 8 < \sqrt{10}$, as they must on either side of the maximum. Range: $f$ increases from values just above $-1$ up to $\sqrt{10}$, then decreases toward $1$, so the range is $(-1, \sqrt{10}]$. The last concave up piece is forced by the asymptote: a decreasing curve that levels off must bend upward.

The cube root exists for every real number, so the domain is $\mathbb{R}$ and $f$ is continuous everywhere: no vertical asymptote. $f(x) = 0 \iff x = 0$ or $x = 5$. $f(-1) = 1 \cdot (-6) = -6$ and $f(1) = -4$: neither even nor odd. Since $x^{2/3} \ge 0$, $f$ has the sign of $x - 5$: negative for $x < 5$ except at $0$, positive after $5$. So the graph TOUCHES the axis at the origin without crossing it. As $x \to \infty$, both factors tend to $\infty$; as $x \to -\infty$, $x^{2/3} \to \infty$ and $x - 5 \to -\infty$, so $f \to -\infty$. No horizontal asymptote, and no slant one either, because $\frac{f(x)}{x} = x^{2/3}\left(1 - \frac{5}{x}\right) \to \infty$: the graph ends up steeper than any line.

The power rule on each term: $f'(x) = \frac{5}{3}x^{2/3} - \frac{10}{3}x^{-1/3}$. Factor out $\frac{5}{3}x^{-1/3}$: $f'(x) = \frac{5}{3}x^{-1/3}(x - 2) = \frac{5(x - 2)}{3x^{1/3}}$. A critical number is a point of the DOMAIN of $f$ where $f' = 0$ or $f'$ does not exist. $f' = 0$ at $x = 2$; $f'$ does not exist at $x = 0$, and $0$ IS in the domain of $f$. So there are two critical numbers, $0$ and $2$. Forgetting $0$ is the classic loss on this chapter: the sign table then runs from $-\infty$ to $2$ in one piece and the local maximum disappears from the sketch.

$x^{1/3}$ has the sign of $x$, so the table of $f'$ reads: for $x < 0$, $\frac{(-)}{(-)} > 0$; for $0 < x < 2$, $\frac{(-)}{(+)} < 0$; for $x > 2$, $f' > 0$. So $f$ increases on $(-\infty, 0]$, decreases on $[0, 2]$ and increases on $[2, \infty)$. First derivative test: local maximum $f(0) = 0$, local minimum $f(2) = 2^{2/3}(2 - 5) = -3 \cdot 2^{2/3} = -3\sqrt[3]{4}$. By hand, $1.5^3 = 3.375 < 4 < 4.096 = 1.6^3$, so $\sqrt[3]{4}$ lies between $1.5$ and $1.6$ and $f(2)$ between $-4.8$ and $-4.5$. At $0$: the numerator tends to $-10$ and $3x^{1/3} \to 0^-$ from the left, $0^+$ from the right, so $f' \to +\infty$ as $x \to 0^-$ and $f' \to -\infty$ as $x \to 0^+$. The tangent lines become vertical from both sides while the graph goes up, then down: a cusp.

$f''(x) = \frac{10}{9}x^{-1/3} + \frac{10}{9}x^{-4/3} = \frac{10}{9}x^{-4/3}(x + 1) = \frac{10(x + 1)}{9x^{4/3}}$. For $x \ne 0$, $x^{4/3} = \left(\sqrt[3]{x}\right)^4 > 0$, so $f''$ has the sign of $x + 1$: concave down on $(-\infty, -1)$, concave up on $(-1, 0)$ and on $(0, \infty)$. At $-1$, which is in the domain, the concavity changes: inflection point $(-1, f(-1)) = (-1, -6)$. At $0$, $f''$ does not exist, but the concavity is up on both sides: the cusp is NOT an inflection point. A missing $f''$ is not a sign change.

The table, left to right: from $-\infty$ the curve rises, concave down, bends at $(-1, -6)$, keeps rising concave up and arrives at the origin with a vertical tangent; it leaves the origin going down, again vertically, falls to $\left(2, -3\sqrt[3]{4}\right)$, then rises through $(5, 0)$ and on to $+\infty$, concave up. A rounded top at the origin would mean $f'(0) = 0$, a horizontal tangent, and the one-sided limits of c) say exactly the opposite. The same limits explain why both sides near the origin are concave up although $0$ is a maximum: the two arcs curve upward, like the two halves of a bird's beak meeting at the tip. That is the picture the computation draws, and the one the marker looks for.

$\ln x$ requires $x > 0$, which also keeps the denominator away from $0$: the domain is $(0, \infty)$. $f(x) = 0 \iff \ln x = 0 \iff x = 1$, and since $x > 0$, $f$ has the sign of $\ln x$: negative on $(0, 1)$, positive on $(1, \infty)$. As $x \to 0^+$, $\ln x \to -\infty$ while $x \to 0^+$: the form is $\frac{-\infty}{0^+}$, which is NOT indeterminate. A very large negative number divided by a very small positive one is very large and negative: $f \to -\infty$, and $x = 0$ is a vertical asymptote, from the right only. A student who applies L'Hospital's rule here writes $\frac{1/x}{1} \to +\infty$, the wrong sign, because the rule needs $\frac{0}{0}$ or $\frac{\infty}{\infty}$ and its conclusion is false outside those forms. As $x \to \infty$, the form is $\frac{\infty}{\infty}$ and the rule applies: $\lim \frac{1/x}{1} = 0$. So $y = 0$ is a horizontal asymptote as $x \to \infty$, approached from above since $f > 0$ there.

Quotient rule: $f'(x) = \frac{\frac{1}{x} \cdot x - \ln x \cdot 1}{x^2} = \frac{1 - \ln x}{x^2}$. Since $x^2 > 0$, $f'$ has the sign of $1 - \ln x$, positive exactly when $\ln x < 1$, that is $x < e$, because $\ln$ is increasing. So $f$ increases on $(0, e]$ and decreases on $[e, \infty)$, and $e$ is the only critical number. The local maximum is the absolute one: $f(e) = \frac{\ln e}{e} = \frac{1}{e}$.

Write $f'(x) = (1 - \ln x)x^{-2}$; product rule: $f''(x) = -\frac{1}{x} \cdot x^{-2} + (1 - \ln x)(-2x^{-3}) = \frac{-1 - 2 + 2\ln x}{x^3} = \frac{2\ln x - 3}{x^3}$. As $x^3 > 0$, $f''$ has the sign of $2\ln x - 3$: negative for $x < e^{3/2}$, positive after. Inflection point at $x = e^{3/2} = e\sqrt e$, with $f\left(e^{3/2}\right) = \frac{3/2}{e^{3/2}} = \frac{3}{2}e^{-3/2}$. The position is coherent with b): after its maximum the curve decreases toward an asymptote from above, so it must end up concave up, and the switch has to happen to the RIGHT of the maximum, which it does since $e^{3/2} > e$. Order of magnitude: $\sqrt e \approx 1.65$ because $1.65^2 \approx 2.72$, so $e^{3/2} \approx 4.5$.

Table: from $x = 0^+$, where $f \to -\infty$, the curve rises, concave down, through $(1, 0)$ to its top $\left(e, \frac{1}{e}\right)$, then decreases, still concave down, until the inflection point near $x = 4.5$, and finally decreases concave up toward $y = 0$. The maximum and the inflection point have close heights, $\frac{1}{e}$ and $\frac{3}{2}e^{-3/2}$, which is why the right part of the graph looks almost flat. Range: on $(0, e]$, $f$ is continuous and increasing from $-\infty$ to $\frac{1}{e}$, so it takes every value of $\left(-\infty, \frac{1}{e}\right]$; on $[e, \infty)$ it takes only values of $\left(0, \frac{1}{e}\right]$. The range is $\left(-\infty, \frac{1}{e}\right]$.

For $x \ne 0$, $g(-x) = \frac{\ln|-x|}{-x} = -g(x)$: $g$ is odd, its graph is symmetric about the origin, and $g = f$ on $(0, \infty)$. The half turn about the origin carries every feature of the right half to the left half, changing the sign of BOTH coordinates. The maximum $\left(e, \frac{1}{e}\right)$ becomes a minimum $\left(-e, -\frac{1}{e}\right)$. The limit $-\infty$ at $0^+$ becomes $+\infty$ at $0^-$: $x = 0$ is a vertical asymptote with opposite behaviours on its two sides. $y = 0$ is an asymptote at both ends, approached from below as $x \to -\infty$. A half turn also swaps concave up and concave down, since $g''(x) = -f''(-x)$: $g$ is concave down on $\left(-\infty, -e^{3/2}\right)$ and concave up on $\left(-e^{3/2}, 0\right)$, and the inflection points are $\left(e^{3/2}, \frac{3}{2}e^{-3/2}\right)$ and $\left(-e^{3/2}, -\frac{3}{2}e^{-3/2}\right)$. At $x = 0$ the concavity changes as well, up on the left and down on the right, but $0$ is not in the domain: no inflection point there. Symmetry is one line of the checklist, and this is what it is worth: half the work for free, provided each feature is transformed, a maximum into a minimum and one concavity into the other.