4.1.30 · D5Calculus I — Limits & Derivatives
Question bank — Curve sketching — systematic approach
Before you start, make sure these are solid in your head:
- Domain = the set of where actually produces a number (no dividing by zero, no square roots of negatives, no logs of non-positive numbers).
- = slope of the tangent line = climb/fall rate. See First derivative and monotonicity.
- = rate of change of the slope = bending. See Second derivative test.
- Asymptote = a line the curve approaches (vertical, horizontal, or oblique). See Rational functions and asymptotes.
- Critical point = where or is undefined.
Two pictures to keep in your mind's eye for the whole page. First, what and look like on a real curve:
Look at the red tangent arrows: where they slope up the function rises (); where they slope down it falls (); at the peak the tangent is flat (). The bending — cup vs cap — is the story layered on top.
Second, the tool almost every "trap" abuses — the sign chart. It compresses all the calculus into a number line of and marks:
A sign chart only carries meaning at points inside the domain and only flags a feature where the sign actually changes. Half the traps below are people reading a sign chart wrong — keep this picture handy.
True or false — justify
Every answer must give the reason, not just the verdict.
Every point where is a local maximum or minimum.
False. must change sign there; e.g. has yet the curve keeps rising through a flat inflection.
If then is an inflection point.
False. must actually change sign at ; e.g. has but stays concave-up on both sides, so no inflection.
A curve can never cross its horizontal asymptote.
False. The horizontal asymptote only controls behaviour as ; crosses infinitely often before settling.
A curve can never cross its vertical asymptote.
True. A vertical asymptote sits at an -value excluded from the domain, so no point of the graph has that -coordinate — nothing to cross.
If is even, it can still have an oblique (slant) asymptote.
False. An even function is symmetric about the -axis, so any slant line it approached on the right forces the mirror line on the left — a single oblique asymptote can't satisfy both, so evenness rules them out (only horizontal asymptotes survive).
A polynomial of degree always has at least one -intercept.
False. Odd-degree polynomials must cross (ends go opposite ways), but even-degree ones like never touch the -axis.
If on an interval, then for any in it.
True. Provided is continuous on and differentiable on , the Mean Value Theorem gives for some ; with and the difference is positive. See Mean Value Theorem.
If is increasing on an interval, then throughout it.
False. Increasing only forces ; is strictly increasing yet .
A function that is concave up everywhere has no local maximum in the interior of its domain.
True. Concave up means the curve lies above its tangents, so any interior critical point is a minimum. Caveat: a maximum can still occur at an endpoint or as a boundary/limiting value — concavity only rules out interior maxima, not boundary ones.
If a rational function has a factor only in its denominator (not cancelled by the numerator), then is a vertical asymptote.
True. A denominator-zero produces a vertical asymptote only when the factor does not cancel; if the same factor also divides the numerator it cancels to a removable hole instead (see the edge-case section), so the "uncancelled" condition is the real rule.
Spot the error
Read the flawed reasoning, then say exactly which step breaks.
"For the asymptote is , so the curve must stay below everywhere."
Wrong. The horizontal asymptote is only the end behaviour. Between the walls () the function is actually negative; outside () it sits above . The line isn't a ceiling.
" is zero at and undefined at , so I have three critical points: ."
Wrong. Critical points must lie in the domain. Here are excluded (asymptotes), so they are not critical points — only is.
" changes sign at for , therefore there are inflection points at ."
Wrong. A genuine inflection point must be a point on the curve. At the function isn't defined, so the concavity flips across a gap, not at an actual point — no inflection there.
" has and , and since is inconclusive I can't classify ."
The Second Derivative Test is inconclusive, but the First Derivative Test finishes the job: on both sides (no sign change), so is a flat inflection, not an extremum.
"To find the oblique asymptote of I take , so the asymptote is ."
You found correctly but stopped early. You still need , which gives , so the asymptote is , not .
"I computed the domain, intercepts, and derivatives, but I got the same peak whether or not I noticed the point isn't in the domain — so the domain step is optional."
Wrong in general. An excluded point can sit exactly where you'd otherwise draw a smooth curve, producing a phantom feature. Skipping the domain step is how hidden holes and asymptotes get missed — it's always step 1.
Why questions
These check that you know why each tool is the right one.
Why do we use the sign of rather than its exact value to decide increasing/decreasing?
Because monotonicity only needs the direction of change; via the Mean Value Theorem the sign of alone fixes the sign of , so magnitude is irrelevant for the trend.
Why does (the derivative of the slope) measure bending rather than steepness?
Steepness is itself; tells you whether that steepness is growing (slope turning upward, concave-up cup) or shrinking (concave-down cap), which is exactly what "bending" means.
Why is the domain checked first in the checklist?
Every later step — intercepts, asymptotes, sign charts — only makes sense on the set where exists; an excluded point can be a hidden asymptote or hole that reshapes the whole picture.
Why does polynomial long division reveal the oblique asymptote?
Division writes where the remainder fraction as , so the curve is squeezed onto the line — the leftover term literally vanishes at the ends.
Why does symmetry ( or ) save work?
You only analyse , then mirror: even functions reflect across the -axis, odd functions rotate about the origin, so half the sign-chart carries over automatically.
Why can't a limit like ever produce a vertical asymptote?
A vertical asymptote is about approaching a finite value while ; the limit at infinity varies , not , so it can only reveal horizontal (or oblique) end behaviour. See Limits and continuity.
Why is a critical point where is undefined still worth checking?
Extrema can occur at sharp corners or cusps (e.g. at ) where the slope doesn't exist; ignoring them would miss real maxima and minima.
Edge cases
The scenarios the checklist quietly assumes away.
What happens at for ?
Numerator and denominator both vanish, so is a removable hole, not an asymptote: simplifies to with a single point punched out at .
Can a function have infinitely many critical points on a bounded interval?
Yes. (with ) oscillates ever faster near , giving critical points accumulating at the origin — the sign chart can't be a finite list there.
What does the graph do at a critical point where changes (no sign change) but ?
It momentarily flattens and then keeps climbing — a stationary inflection (like at ), which is neither a max nor a min.
If both and , is the point automatically an inflection?
No. Both being zero is inconclusive; you must test sign changes. at has yet is a genuine minimum, while at is an inflection.
Does a horizontal asymptote at have to equal the one at ?
No. A function can approach different horizontal levels on each side, e.g. tends to as and as — two separate horizontal asymptotes.
What is the concavity of a straight line, and does it have inflection points?
A line has everywhere, so it is neither concave up nor down and has no inflection points — concavity requires actual bending, which a line never does.
Can a function be both even and odd?
Only the zero function . Evenness gives and oddness gives ; together they force , hence for all .
At a vertical asymptote, must the curve go to on both sides?
No. The two one-sided limits can differ in sign — e.g. dives to from the left of and shoots to from the right; you must check each side separately.
Recall One-line summary of the trap pattern
Almost every trap here is one of three shapes: (1) confusing a necessary condition (, ) with a sufficient one — always demand a sign change; (2) treating an asymptote as a barrier instead of a limiting behaviour; (3) forgetting the domain, which turns "asymptotes" into holes and phantom critical points on and off. Reason, don't reflex.