Curve Sketching and Optimization

Monotonicity, extrema, concavity, and end behavior are separate readings of a function. A hand sketch merges them into one picture. Optimization runs the same extremum tests in the other direction: reduce a real quantity to a function of one variable and find its extreme value.

The curve-sketching checklist

A calculus sketch places the features that matter — intercepts, asymptotes, extrema, inflections — rather than sampling points blindly. Stewart organizes the work as eight steps.¹

• A. Domain. Determine the set $D$ of $x$ for which $f(x)$ is defined.
• B. Intercepts. The $y$-intercept is $f(0)$. Find $x$-intercepts by solving $f(x)=0$ (skip if intractable).
• C. Symmetry. If $f(-x)=f(x)$, $f$ is even and the graph is symmetric about the $y$-axis. If $f(-x)=-f(x)$, $f$ is odd and symmetric about the origin. If $f(x+p)=f(x)$, $f$ is periodic with period $p$.
• D. Asymptotes. Horizontal: ${\lim }_{x\to \pm \infty }f(x)=L$ gives $y=L$. Vertical: $x=a$ where a one-sided limit is $\pm \infty$. Slant: handled below.
• E. Intervals of increase/decrease. Sign of $f^{\prime }$ by the Increasing/Decreasing Test.
• F. Local extrema. Critical numbers plus the First (or Second) Derivative Test.
• G. Concavity and inflection. Sign of $f^{\prime \prime }$ by the Concavity Test; inflections where it changes.
• H. Sketch. Draw asymptotes as dashed lines, plot the intercepts and the extreme and inflection points, then join them respecting E and G.

Symmetry is a labor saver. For an even or odd function, sketch $x\ge 0$ and reflect; for a periodic function, sketch one period and translate.

A worked sketch

Apply the checklist to $y=\frac{2x^2}{x^2-1}$.

• Domain: all $x$ with $x^2\ne 1$, so $x\ne \pm 1$.
• Intercepts: $f(0)=0$; the only $x$-intercept is the origin.
• Symmetry: $f(-x)=f(x)$, so the graph is even.
• Asymptotes: ${\lim }_{x\to \pm \infty }\frac{2x^2}{x^2-1}=2$, so $y=2$ is a horizontal asymptote. The denominator vanishes at $x=\pm 1$, giving vertical asymptotes there.
• Monotonicity: $f^{\prime }(x)=\frac{-4x}{(x^2-1{)}^2}$, positive for $x<0$ and negative for $x>0$ (away from $\pm 1$). So $f$ increases on $(-\infty ,-1)$ and $(-1,0)$, decreases on $(0,1)$ and $(1,\infty )$.
• Extrema: $f^{\prime }(0)=0$ with a sign change from $+$ to $-$, so $f(0)=0$ is a local maximum.
• Concavity: $f^{\prime \prime }(x)=\frac{12x^2+4}{(x^2-1{)}^3}$, positive when $x^2>1$ and negative when $x^2<1$, so the graph is concave up on $(-\infty ,-1)$ and $(1,\infty )$ and concave down on $(-1,1)$. No inflection points, since $x=\pm 1$ are not in the domain.

Slant asymptotes

When the numerator of a rational function has degree exactly one more than the denominator, the graph approaches a slanted line rather than a horizontal one. Polynomial division writes

so $f(x)-(mx+b)\to 0$ and the line $y=mx+b$ is a slant asymptote. The curve approaches that line at both ends, above it where the remainder is positive and below where it is negative.

Optimization: modeling one quantity

The extremum tools answer applied questions once the situation is written as a function. The difficulty is almost always the setup, not the calculus. A fixed template turns a word problem into a single-variable extremum.

The last step matters: a problem asking for dimensions is not answered by the critical number alone.

Maximizing an area under a constraint

A farmer has $2400$ ft of fencing for a rectangular field bordering a straight river, needing no fence along the river. Which dimensions maximize the area?

Let $x$ be the depth (the two sides perpendicular to the river) and $y$ the width (the side parallel to it). The area is $A=xy$, and the fence covers three sides:

Substituting eliminates $y$:

Then $A^{\prime }(x)=2400-4x=0$ at $x=600$. Since $A^{\prime \prime }(x)=-4<0$, the graph is concave down everywhere and the critical number is the absolute maximum; $A(0)=A(1200)=0$ confirms it by the Closed Interval Method. The width is $y=2400-2(600)=1200$, so the field is $600$ ft deep and $1200$ ft wide, with area $720,000$ ft${}^2$.

Minimizing material with an open domain

A cylindrical can must hold $1$ L $=1000$ cm${}^3$. Which radius and height use the least metal? The cost tracks the total surface area of the top, bottom, and side.

The side unrolls into a rectangle of height $h$ and width equal to the circumference $2\pi r$, so its area is $2\pi rh$; the two disks contribute $2\pi r^2$. Thus $A=2\pi r^2+2\pi rh$. The volume constraint $\pi r^{2}h=1000$ gives $h=1000/(\pi r^2)$, and substituting eliminates $h$:

The domain is open, $r>0$, so the Closed Interval Method does not apply and the First Derivative Test for absolute extrema takes over. Differentiate:

The only critical number is $r=\sqrt[3]{500/\pi }\approx 5.42$ cm. For $r$ below it $A^{\prime }<0$ and above it $A^{\prime }>0$, so this critical number is the absolute minimum.

The corresponding height is $h=1000/(\pi r^2)=2r$: the most economical can is exactly as tall as it is wide across the diameter. This single-critical-number argument is the standard replacement for the Closed Interval Method whenever the physical domain is an open or infinite interval.

The two examples used different closing arguments because their domains differed. The choice is dictated by whether the domain is a closed bounded interval.

Minimizing distance to a curve

A closest-point problem models the distance from a fixed point to a curve, then minimizes it. Minimizing the squared distance avoids the square root and has the same minimizer.

Least-time paths

Fermat's principle in optics states that light takes the path of least time. The same structure governs a traveler choosing where to cross terrain of two speeds.