Corrected exercises: function analysis and curve sketching (201-SN2-RE)

Domain $\mathbb{R}\setminus\{-2\}$, and at $x=-2$ the numerator equals $-7\neq 0$: vertical asymptote $x=-2$. Equal degrees: $\lim_{x\to\pm\infty}f(x)=\frac{3}{1}$, horizontal asymptote $y=3$, the ratio of the leading coefficients.

$x^{2}-2x-15=(x+3)(x-5)$; at $x=-3$ the numerator equals $1$, at $x=5$ it equals $9$: two vertical asymptotes, $x=-3$ and $x=5$. Since the degree of the numerator is lower, the horizontal asymptote is $y=0$. Pitfall: looking for the zeros of the NUMERATOR, $x=-4$, which give a zero of $f$ and not an asymptote.

$f(x)=\frac{(x-2)(x+2)}{(x-2)(x+3)}=\frac{x+2}{x+3}$ for $x\neq 2$. At $x=2$, the factor cancels: the curve has a HOLE at the point $\left(2, \frac{4}{5}\right)$, not an asymptote. Only $x=-3$ is a vertical asymptote; the horizontal one is $y=1$. Pitfall: declaring $x=2$ an asymptote because the denominator vanishes there.

Long division: $2x^{2}-3x+4=(x+1)(2x-5)+9$, so $f(x)=2x-5+\frac{9}{x+1}$. The remainder tends to $0$: oblique asymptote $y=2x-5$, with $m=2$ and $b=-5$ (and vertical asymptote $x=-1$). Pitfall: stopping at $2x$, the first term of the quotient.

Division by $x^{2}+1$: $x^{3}-2x^{2}+5=(x^{2}+1)(x-2)+(-x+7)$, so $f(x)=x-2+\frac{-x+7}{x^{2}+1}$. The remainder is degree $1$ over degree $2$: it tends to $0$. Oblique asymptote $y=x-2$, $m=1$, $b=-2$, and no vertical asymptote since $x^{2}+1$ never vanishes.

The numerator exceeds the denominator by TWO degrees: $f(x)=x^{2}+2x+4+\frac{9}{x-2}$, which behaves like a parabola. No horizontal or oblique asymptote. At $x=2$ the numerator equals $9\neq 0$: vertical asymptote $x=2$. Pitfall: taking the quotient $x^{2}+2x+4$ for an oblique asymptote, when an oblique asymptote is a LINE.

For $x\neq 0$, $\sqrt{4x^{2}+5}=|x|\sqrt{4+\frac{5}{x^{2}}}$. At $+\infty$, $|x|=x$ and $f(x)=\frac{3-\frac{1}{x}}{\sqrt{4+\frac{5}{x^{2}}}}\to\frac{3}{2}$. At $-\infty$, $|x|=-x$ and $f(x)\to-\frac{3}{2}$. Two horizontal asymptotes, $y=\frac{3}{2}$ and $y=-\frac{3}{2}$. Pitfall: writing $\sqrt{x^{2}}=x$ on both sides.

At $+\infty$, $e^{-2x}\to 0$ and $f(x)\to 4$: horizontal asymptote $y=4$. At $-\infty$, $e^{-2x}\to+\infty$ and $f(x)\to-\infty$: nothing on that side. Domain $\mathbb{R}$, no vertical asymptote. The asymptote therefore exists at $+\infty$ only.

Domain: $x+4>0$, that is $x>-4$. As $x\to-4^{+}$, $x+4\to 0^{+}$, $\ln(x+4)\to-\infty$, so $f(x)=1-\ln(x+4)\to+\infty$. Vertical asymptote $x=-4$, with the curve going UP. Pitfall: the minus sign in front of the logarithm flips the limit.

At $-\infty$, $e^{x}\to 0$, so $f(x)-\left(\frac{x}{2}+3\right)=e^{x}\to 0$: oblique asymptote $y=\frac{x}{2}+3$, with $m=\frac{1}{2}$ and $b=3$, at $-\infty$ only. At $+\infty$, the gap $e^{x}$ blows up: no asymptote on that side. Pitfall: looking for it at $+\infty$ out of habit.

$f'(x)=6x-12$, zero at $x=2$; $f''(x)=6>0$, so a minimum, with value $f(2)=12-24+7=-5$.

$f'(x)=6x^{2}+6x-12=6(x+2)(x-1)$: critical numbers $-2$ and $1$. $f''(x)=12x+6$: $f''(-2)=-18<0$, local maximum at $x=-2$; $f''(1)=18>0$, minimum at $x=1$. $f''$ vanishes and changes sign at $x=-\frac{1}{2}$: inflection point. Pitfall: taking the first zero found for the maximum without looking at the sign of $f''$.

$f'(x)=3x^{2}+6x+6=3\left((x+1)^{2}+1\right)>0$: discriminant $36-72<0$, NO critical number, the function is strictly increasing. $f''(x)=6x+6$ changes sign at $x=-1$: inflection point. Pitfall: believing that a cubic always has two critical numbers.

$f'(x)=4x^{3}-16x=4x(x-2)(x+2)$: three critical numbers, $-2$, $0$ and $2$. $f''(x)=12x^{2}-16$, zero at $x=\pm\frac{2}{\sqrt{3}}$, with a sign change: the inflection point on the positive side is at $x=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\approx 1.155$. Pitfall: dividing by $x$ and losing the critical number $0$.

$f'(x)=4x^{3}-12x^{2}+12x=4x(x^{2}-3x+3)$, and $x^{2}-3x+3$ has discriminant $9-12<0$: the only critical number is $x=0$ (a minimum, $f(0)=0$). $f''(x)=12x^{2}-24x+12=12(x-1)^{2}$: zero at $x=1$, but POSITIVE on both sides. No sign change, so NO inflection point at $x=1$; the curve stays concave up everywhere.

Domain $\mathbb{R}\setminus\{0\}$. $f'(x)=1-\frac{9}{x^{2}}=\frac{(x-3)(x+3)}{x^{2}}$: critical numbers $\pm 3$, the positive one being $3$ (local minimum, $f(3)=6$). $f''(x)=\frac{18}{x^{3}}$ never vanishes; it changes sign at $0$, but $0$ is not in the domain. No inflection point. Pitfall: counting $x=0$.

$f'(x)=-\frac{12x}{(x^{2}+3)^{2}}$, zero at $x=0$, positive before and negative after: maximum $f(0)=2$. $f''(x)=\frac{36x^{2}-36}{(x^{2}+3)^{3}}=\frac{36(x-1)(x+1)}{(x^{2}+3)^{3}}$, which changes sign at $\pm 1$: inflection points at $x=-1$ and $x=1$. Pitfall: in the derivative of the quotient, forgetting the factor $2\cdot 2x$ that the chain rule gives on $(x^{2}+3)^{2}$.

Domain $[0, +\infty)$. $f(x)=x^{\frac{3}{2}}-3x^{\frac{1}{2}}$, so $f'(x)=\frac{3}{2}\sqrt{x}-\frac{3}{2\sqrt{x}}=\frac{3(x-1)}{2\sqrt{x}}$. Two critical numbers: $x=1$, where $f'$ vanishes, and $x=0$, in the domain, where $f'$ does not exist. Minimum at $x=1$: $f(1)=1\times(-2)=-2$. $f''(x)=\frac{3(x+1)}{4x\sqrt{x}}>0$: no inflection point. Pitfall: forgetting $x=0$, which is critical because $f'$ does not exist there.

$f'(x)=2xe^{x}+x^{2}e^{x}=x(x+2)e^{x}$: critical numbers $0$ and $-2$, the negative one being $-2$ (local maximum). $f''(x)=(x^{2}+4x+2)e^{x}$, zero at $x=-2\pm\sqrt{2}$ with a sign change. The one closest to $0$: $x=-2+\sqrt{2}\approx-0.586$. Pitfall: looking for a zero of $e^{x}$, which has none.

$f'(x)=2\cos x-2\sin(2x)=2\cos x-4\sin x\cos x=2\cos x\,(1-2\sin x)$. $\cos x=0$ gives $\frac{\pi}{2}$ and $\frac{3\pi}{2}$, $\sin x=\frac{1}{2}$ gives $\frac{\pi}{6}$ and $\frac{5\pi}{6}$: four critical numbers. Values: $f\left(\frac{\pi}{6}\right)=f\left(\frac{5\pi}{6}\right)=1+\frac{1}{2}=\frac{3}{2}$, $f\left(\frac{\pi}{2}\right)=1$, $f\left(\frac{3\pi}{2}\right)=-3$, and at the endpoints $f(0)=f(2\pi)=1$. Maximum $\frac{3}{2}$. Pitfall: dividing by $\cos x$ and keeping only the two solutions of $\sin x=\frac{1}{2}$.

Domain: the denominator $x^{2}-9=(x-3)(x+3)$ vanishes at $x=\pm 3$, so $D_{f}=\mathbb{R}\setminus\{-3,3\}$. These values do not cancel with the numerator $2x^{2}$, so they really are VERTICAL asymptotes: $x=3$ and $x=-3$. Horizontal asymptote: $\lim_{x\to\pm\infty}\frac{2x^{2}}{x^{2}-9}=\frac{2}{1}=2$ (equal degrees, ratio of the leading coefficients), so $y=2$. Checking the behaviour near $x=3$: on the right the denominator is positive and small, so $f\to+\infty$; on the left it is negative, so $f\to-\infty$. This sign detail is read from the table and it is required on the exam.

The degree of the numerator (2) exceeds that of the denominator (1) by exactly 1: there is therefore an oblique asymptote. Long division: $x^{2}+1=(x-1)(x+1)+2$, hence $g(x)=x+1+\frac{2}{x-1}$. Since $\frac{2}{x-1}\to 0$ as $x\to\pm\infty$, the curve hugs the line $y=x+1$: it is the oblique asymptote. As a bonus, the remainder gives the SIDE of approach: for $x>1$ the remainder is positive and the curve lies above the asymptote; for $x<1$ it is negative and the curve lies below. There is also a vertical asymptote at $x=1$.

Let $d_{N}$ be the degree of the numerator and $d_{D}$ that of the denominator. If $d_{N}<d_{D}$: horizontal asymptote $y=0$, for example $\frac{1}{x^{2}+1}$. If $d_{N}=d_{D}$: horizontal asymptote $y=$ ratio of the leading coefficients, for example $\frac{2x^{2}}{x^{2}-9}$ with $y=2$ as in part a). If $d_{N}=d_{D}+1$: OBLIQUE asymptote obtained by division, as in part b). If $d_{N}\geq d_{D}+2$: neither horizontal nor oblique, the curve goes off to infinity following a parabola or a curve of higher degree, for example $\frac{x^{3}}{x-1}$, which behaves like $x^{2}+x+1$. The criterion is purely a matter of degrees and is checked in three seconds.

Counterexample: $h(x)=\frac{2x}{x^{2}+1}$ has the horizontal asymptote $y=0$ (the degree of the numerator is lower), and yet $h(0)=0$: the curve CROSSES its asymptote at the origin, before approaching it from above on one side and from below on the other. The claim is therefore false in general: an asymptote describes a behaviour at INFINITY, it imposes nothing at finite distance. It is true, however, for VERTICAL asymptotes, which no curve can cross: the value $x=a$ is excluded from the domain, so the curve has no point with $x$-coordinate $a$. The distinction is exactly the one between a domain constraint, which is absolute, and a limit constraint, which is only asymptotic.

$f'(x)=3x^{2}-6x-9=3(x^{2}-2x-3)=3(x-3)(x+1)$, which vanishes at $x=-1$ and $x=3$. Signs: for $x<-1$ both factors are negative, so $f'>0$; between $-1$ and 3, $(x+1)>0$ and $(x-3)<0$, so $f'<0$; for $x>3$, $f'>0$. The function increases on $\left(-\infty, -1\right]$, decreases on $\left[-1, 3\right]$, increases on $\left[3, +\infty\right)$. At $x=-1$, $f'$ changes from $+$ to $-$: local MAXIMUM, with value $f(-1)=-1-3+9+5=10$. At $x=3$, $f'$ changes from $-$ to $+$: local MINIMUM, with value $f(3)=27-27-27+5=-22$.

$f''(x)=6x-6=6(x-1)$. At $x=-1$: $f''(-1)=-12<0$, so local maximum ✓. At $x=3$: $f''(3)=12>0$, so local minimum ✓. The two tests agree, as they always must when the second one applies. The second derivative test is faster when it works, since it takes a single evaluation instead of a sign study over a whole interval; but it says nothing about the GLOBAL behaviour, which only the sign chart of $f'$ provides. On the exam, the table therefore remains mandatory.

$g(x)=x^{4}$ gives $g'(x)=4x^{3}$, which vanishes at $x=0$: it is indeed a critical number. Then $g''(x)=12x^{2}$, so $g''(0)=0$: the second derivative test does NOT conclude, it is inconclusive. Go back to the first derivative test: $g'(x)=4x^{3}$ is negative for $x<0$ and positive for $x>0$, so $g'$ changes from $-$ to $+$ and $x=0$ is a local MINIMUM (and even an absolute one). The example shows that $f''(c)=0$ allows no conclusion, in either direction: you must then go back down one level.

The standard example is $f(x)=x^{3}$ at $x=0$: $f'(x)=3x^{2}$ does vanish at 0, which is therefore a critical number, but $f'(x)>0$ on both sides, so $f'$ does NOT CHANGE sign. There is neither a maximum nor a minimum: the function keeps increasing, only pausing with a horizontal tangent on the way. The point is, however, an inflection point, since $f''(x)=6x$ changes sign at 0. The conclusion to remember: “critical number” and “extremum” are not synonyms. A critical number is only a CANDIDATE, and it is the sign change of $f'$ that decides. Many students lose marks by announcing an extremum as soon as they have set the derivative to zero, without drawing up the sign chart.