Exercises — Curve sketching — systematic approach

Domain. The only thing that can break a fraction is dividing by zero. Set the denominator to zero: $x-3=0\Rightarrow x=3$. So the function is defined for every real number except $3$: $\text{Domain}=\mathbb{R}\setminus \{3\}.$ Vertical asymptote. At $x=3$ the denominator $\to 0$ while the numerator $\to 3+2=5\ne 0$. A nonzero number divided by something tiny explodes, so ${\lim }_{x\to 3^{\pm }}f(x)=\pm \infty \Rightarrow \text{vertical asymptote at }x=3.$ Why numerator $\ne 0$ matters: if both top and bottom vanished, we might have a removable hole instead of a wall — always check the numerator. See Rational functions and asymptotes.

Replace every $x$ by $-x$ and simplify: $g(-x)=(-x{)}^4-3(-x{)}^2+1=x^4-3x^2+1=g(x).$ Because $(-x{)}^{\text{even power}}=x^{\text{even power}}$, nothing changed. Since $g(-x)=g(x)$, the function is even — its graph is a mirror image across the $y$-axis. (Only even powers of $x$ appear, which is the quick fingerprint of an even polynomial.)

A horizontal asymptote asks: where does the curve settle as $x$ runs off to infinity? Divide top and bottom by the highest power, $x^2$: $h(x)=\frac{3+\frac{1}{x^2}}{1+\frac{5}{x^2}}.$ As $x\to \pm \infty$, the terms $\frac{1}{x^2}$ and $\frac{5}{x^2}$ shrink to $0$ (a fixed number over a huge number). So ${\lim }_{x\to \pm \infty }h(x)=\frac{3+0}{1+0}=3\Rightarrow y=3.$ This uses the limit idea from Limits and continuity.

$y$-intercept: plug $x=0$: $f(0)=0-0=0$. Point $(0,0)$. $x$-intercepts: solve $f(x)=0$. Factor first — pull out the common $x$: $x^3-4x=x(x^2-4)=x(x-2)(x+2)=0.$ A product is zero exactly when one factor is zero, so $x=0,2,-2$. Intercepts at $(-2,0),(0,0),(2,0)$.

Differentiate. $f^{\prime }(x)=3x^2-6x=3x(x-2).$ Critical points are where $f^{\prime }=0$ (here $f^{\prime }$ is never undefined): $3x(x-2)=0\Rightarrow x=0,2$. Sign chart of $f^{\prime }$ (test one number in each region — see First derivative and monotonicity):

• $x<0$, say $x=-1$: $3(-1)(-3)=+9>0$ → rising.
• $0<x<2$, say $x=1$: $3(1)(-1)=-3<0$ → falling.
• $x>2$, say $x=3$: $3(3)(1)=+9>0$ → rising.

At $x=0$: $+\to -$ ⇒ local maximum, value $f(0)=0$ → $(0,0)$. At $x=2$: $-\to +$ ⇒ local minimum, value $f(2)=8-12=-4$ → $(2,-4)$.

From above, $f^{\prime }(x)=3x^2-6x$. Differentiate again: $f^{\prime \prime }(x)=6x-6=6(x-1).$ $f^{\prime \prime }$ measures whether the slope is itself rising or falling (Second derivative test).

• $x<1$: $f^{\prime \prime }<0$ → concave down ($⌢$).
• $x>1$: $f^{\prime \prime }>0$ → concave up ($⌣$).

The concavity flips at $x=1$, and $f^{\prime \prime }$ changes sign there, so $(1,f(1))=(1,1-3)=(1,-2)$ is an inflection point.

$f^{\prime }(x)=3x^2$, which is $0$ at $x=0$ — a genuine critical point. Now test the sign of $f^{\prime }$ on both sides:

• $x<0$: $3x^2>0$ (a square is positive).
• $x>0$: $3x^2>0$.

The slope is positive on both sides — it never changes sign — so by the First Derivative Test there is no local extremum. Instead check $f^{\prime \prime }(x)=6x$: it is $-$ for $x<0$, $+$ for $x>0$, so concavity flips at $0$. That makes $(0,0)$ a flat (horizontal) inflection point: the tangent is level, yet the curve keeps climbing while switching from $⌢$ to $⌣$.

What the figure shows: the blue curve $x^3$ is drawn through the origin; the yellow dashed segment is the horizontal tangent (slope $0$) touching at $(0,0)$. To the left of the origin the curve is labelled concave down ($⌢$, red), to the right concave up ($⌣$, green). Notice that despite the flat tangent, the curve keeps rising left-to-right — the visual proof that a zero slope need not be a peak or valley.

The numerator's degree ($2$) is exactly one more than the denominator's ($1$), the signal for a slant asymptote. Do the division: $f(x)=\frac{x^2+1}{x}=x+\frac{1}{x}.$ The leftover $\frac{1}{x}\to 0$ as $x\to \pm \infty$, so the curve hugs the line $y=x.$ How it approaches: for large positive $x$, $\frac{1}{x}>0$, so the curve sits just above $y=x$. For large negative $x$, $\frac{1}{x}<0$, so it sits just below. It squeezes toward the line from opposite sides — never touching, since $\frac{1}{x}\ne 0$.

Domain. Denominator $x^2+1$ is always $\ge 1>0$ (a square plus one can never be zero). So $f$ is defined for all real $x$: domain $=\mathbb{R}$. No vertical asymptote: vertical asymptotes need the denominator to hit $0$; it never does, so there are none. This is the payoff of checking the domain first. Symmetry. $f(-x)=\frac{(-x{)}^2}{(-x{)}^2+1}=\frac{x^2}{x^2+1}=f(x)$ → even. Horizontal asymptote. Divide by $x^2$: $f(x)=\frac{1}{1+1/x^2}\to \frac{1}{1+0}=1$ as $x\to \pm \infty$, so $y=1$. Since $x^2<x^2+1$ always, $f(x)<1$ everywhere — the curve approaches $y=1$ from below and never reaches it.

1. Domain. $x^2-4=(x-2)(x+2)=0$ at $x=\pm 2$: domain $\mathbb{R}\setminus \{-2,2\}$. 2. Intercepts. $f(0)=\frac{1}{-4}=-\frac{1}{4}$ → $(0,-\frac{1}{4})$. Set $f=0$: a fraction is zero only when its numerator is zero, but the numerator is the constant $1\ne 0$, so no $x$-intercept. 3. Symmetry. $f(-x)=\frac{1}{x^2-4}=f(x)$ → even, mirror over $y$-axis. 4. Asymptotes. Vertical at $x=\pm 2$ (denominator $\to 0$, numerator $=1\ne 0$). Horizontal: $\frac{1}{x^2-4}\to 0$ as $x\to \pm \infty$ (constant over huge number), so $y=0$. 5. First derivative. Write $f=(x^2-4{)}^{-1}$ and use the chain rule: $f^{\prime }(x)=-1\cdot (x^2-4{)}^{-2}\cdot 2x=\frac{-2x}{(x^2-4{)}^2}.$ Denominator $>0$ everywhere it's defined. Numerator $-2x=0$ at $x=0$. Sign: $f^{\prime }>0$ for $x<0$, $f^{\prime }<0$ for $x>0$ ⇒ local maximum at $(0,-\frac{1}{4})$. 6. Second derivative and concavity. Differentiate $f^{\prime }=-2x(x^2-4{)}^{-2}$ with the product + chain rules: $f^{\prime \prime }(x)=-2(x^2-4{)}^{-2}+(-2x)\cdot (-2)(x^2-4{)}^{-3}\cdot 2x=\frac{-2(x^2-4)+8x^2}{(x^2-4{)}^3}=\frac{6x^2+8}{(x^2-4{)}^3}.$ The numerator $6x^2+8$ is always positive (sum of a square-term and $8$). So the sign of $f^{\prime \prime }$ is the sign of the denominator $(x^2-4{)}^3$:

• $|x|>2$: $x^2-4>0\Rightarrow (x^2-4{)}^3>0\Rightarrow f^{\prime \prime }>0$ → concave up ($⌣$).
• $|x|<2$: $x^2-4<0\Rightarrow (x^2-4{)}^3<0\Rightarrow f^{\prime \prime }<0$ → concave down ($⌢$).

Why the sign flips: the numerator never changes sign, so only the denominator can flip $f^{\prime \prime }$, and it flips exactly where $x^2-4$ crosses zero — at $x=\pm 2$. But those points are not in the domain, so there is no inflection point: the concavity changes across a wall, not at a point on the curve. 7. Assemble. Three pieces: a left branch (for $x<-2$) diving from $+\infty$ toward $0$ and concave up; a middle hump (concave down) peaking at $(0,-\frac{1}{4})$ and falling to $-\infty$ at both walls; a right branch (for $x>2$) mirroring the left, concave up.

What the figure shows: the three blue branches of $1/(x^2-4)$; the two red dashed vertical lines are the walls $x=\pm 2$; the yellow dashed horizontal line is the floor $y=0$. The green dot marks the local maximum $(0,-\frac{1}{4})$ on the middle hump — note it lies below the $x$-axis yet is still the highest point of its branch. Watch each outer branch flatten toward $y=0$ (concave up) while the middle branch caps downward.

Domain: all $\mathbb{R}$ (the exponential is defined everywhere). Intercepts: $f(0)=0\cdot e^0=0$. $f(x)=0$ needs $x=0$ (since $e^{-x}$ is never $0$). Only $(0,0)$. First derivative (product rule): $f^{\prime }(x)=1\cdot e^{-x}+x\cdot (-e^{-x})=e^{-x}(1-x).$ $e^{-x}>0$ always, so the sign is that of $(1-x)$: $f^{\prime }>0$ for $x<1$, $f^{\prime }<0$ for $x>1$ ⇒ $+\to -$ ⇒ local maximum at $x=1$, height $f(1)=1\cdot e^{-1}=\frac{1}{e}\approx 0.368$. Second derivative: $f^{\prime \prime }(x)=\frac{d}{dx}[e^{-x}(1-x)]=-e^{-x}(1-x)+e^{-x}(-1)=e^{-x}(x-2).$ Sign is that of $(x-2)$: concave down for $x<2$, concave up for $x>2$ ⇒ inflection at $x=2$, height $f(2)=2e^{-2}\approx 0.271$. End behaviour. As $x\to +\infty$, $e^{-x}$ decays far faster than $x$ grows, so $f\to 0^{+}$ (approaches the $x$-axis from above). As $x\to -\infty$, $x<0$ and $e^{-x}\to +\infty$, so $f\to -\infty$. Shape: rises from $-\infty$, crosses origin, peaks at $(1,\frac{1}{e})$, bends over through the inflection at $x=2$, then decays toward $y=0$.

Domain: $x-2=0$ at $x=2$: $\mathbb{R}\setminus \{2\}$. Vertical asymptote at $x=2$ (numerator there is $4-3=1\ne 0$). Intercepts: $f(0)=\frac{-3}{-2}=\frac{3}{2}$. $f=0$ needs $x^2-3=0\Rightarrow x=\pm \sqrt{3}\approx \pm 1.732$. Oblique asymptote. Degree of top ($2$) is one more than bottom ($1$) → slant line. Polynomial division: $\frac{x^2-3}{x-2}=x+2+\frac{1}{x-2}.$ (Check: $(x+2)(x-2)=x^2-4$, plus remainder $1$ gives $x^2-3$. ✓) The remainder $\frac{1}{x-2}\to 0$ at infinity, so the asymptote is $y=x+2$. Above it for $x>2$, below it for $x<2$. First derivative. Differentiate the divided form (easiest): $f=x+2+(x-2{)}^{-1}$, so $f^{\prime }(x)=1-(x-2{)}^{-2}=1-\frac{1}{(x-2{)}^2}.$ Set $f^{\prime }=0$: $(x-2{)}^2=1\Rightarrow x-2=\pm 1\Rightarrow x=1$ or $x=3$.

• $f(1)=1+2+\frac{1}{-1}=2$ → $(1,2)$.
• $f(3)=3+2+\frac{1}{1}=6$ → $(3,6)$.

Classify with the First Derivative Test (sign of $f^{\prime }(x)=1-\frac{1}{(x-2{)}^2}$). Notice the term $\frac{1}{(x-2{)}^2}$ is large when $x$ is close to $2$ and shrinks toward $0$ when $x$ is far from $2$; so $f^{\prime }>0$ far from the wall and $f^{\prime }<0$ near it. Test one point in each region:

• $x<1$, say $x=0$: $f^{\prime }=1-\frac{1}{4}=+\frac{3}{4}>0$ → rising.
• $1<x<2$, say $x=1.5$: $f^{\prime }=1-\frac{1}{0.25}=1-4=-3<0$ → falling.
• $2<x<3$, say $x=2.5$: $f^{\prime }=1-\frac{1}{0.25}=1-4=-3<0$ → falling.
• $x>3$, say $x=4$: $f^{\prime }=1-\frac{1}{4}=+\frac{3}{4}>0$ → rising.

Applying the test:

• At $x=1$: $f^{\prime }$ changes $+\to -$ ⇒ local maximum at $(1,2)$.
• At $x=3$: $f^{\prime }$ changes $-\to +$ ⇒ local minimum at $(3,6)$.

Describe both branches (the wall $x=2$ splits the curve in two):

• Left branch ($x<2$): comes up from $-\infty$ along the asymptote $y=x+2$ far to the left, rises to the local maximum $(1,2)$, then turns and plunges to $-\infty$ as $x\to 2^{-}$ (denominator $\to 0^{-}$, so $\frac{1}{x-2}\to -\infty$). This whole branch sits below the line $y=x+2$ because the remainder $\frac{1}{x-2}<0$ there.
• Right branch ($x>2$): descends from $+\infty$ as $x\to 2^{+}$ (denominator $\to 0^{+}$, so $\frac{1}{x-2}\to +\infty$), reaches the local minimum $(3,6)$, then climbs and hugs the asymptote $y=x+2$ from above as $x\to +\infty$ (remainder $\frac{1}{x-2}>0$).

Curiosity worth noting: the local minimum $(3,6)$ sits higher than the local maximum $(1,2)$. That is perfectly legal because the two extrema live on different branches separated by the vertical wall — you never travel continuously from one to the other.

General argument. By definition an inflection point is a point where $f^{\prime \prime }$ changes sign. We are told $f^{\prime \prime }(0)=0$ and that $f^{\prime \prime }$ changes sign at $0$ — that is exactly the definition satisfied — so $(0,f(0))$ is an inflection point. Since the function passes through the origin, $f(0)=0$, so the inflection is precisely at $(0,0)$. (Oddness guarantees $f(0)=0$ automatically, because $f(-0)=-f(0)\Rightarrow f(0)=0$.) Concrete check with $f(x)=x^3-3x$:

• Odd? $f(-x)=(-x{)}^3-3(-x)=-x^3+3x=-(x^3-3x)=-f(x)$ ✓.
• Through origin? $f(0)=0^3-3\cdot 0=0$ ✓.
• $f^{\prime \prime }(0)=0$? First $f^{\prime }(x)=3x^2-3$, then $f^{\prime \prime }(x)=6x$, so $f^{\prime \prime }(0)=6\cdot 0=0$ ✓.
• Sign change of $f^{\prime \prime }$ at $0$? $f^{\prime \prime }(-1)=6(-1)=-6<0$ (concave down just left of $0$) and $f^{\prime \prime }(1)=6(1)=+6>0$ (concave up just right of $0$) — the sign flips $-\to +$ ✓.

All four hypotheses hold, so $(0,0)$ is an inflection point of $x^3-3x$, matching the parent note's Worked Example 2. This ties together Second derivative test and symmetry.

Let the two sides perpendicular to the wall each be $x$, and the side parallel be $y$. Fencing used: $2x+y=60\Rightarrow y=60-2x$. Area: $A(x)=xy=x(60-2x)=60x-2x^2,0<x<30.$ (Domain: $x>0$ for a real side, and $y=60-2x>0$ forces $x<30$.) First derivative: $A^{\prime }(x)=60-4x$. Set $=0$: $x=15$. Classify with second derivative: $A^{\prime \prime }(x)=-4<0$ everywhere, so the curve is concave down (a $⌢$), confirming $x=15$ is a maximum (this is the Second derivative test; connects to Optimization). Dimensions: $x=15\text{ m}$, $y=60-2(15)=30\text{ m}$. Maximum area $=15\times 30=450{\text{ m}}^2$.