Dawson 201-SN2-RE: differential calculus tutoring

Curve-analysis workshop: when the asymptote is a line

This original Dawson 201-SN2-RE practice extends the existing curve-sketching lesson with a slant asymptote and a derivative-graph transfer. It is a topic-based study option, not a prediction of your exam. Check your section’s outline before including curve applications or slant asymptotes in a timed review.

If your next assessment covers only early topics: work through domains and limits, then choosing derivative rules. Return here when your outline includes applications. For the earlier three-branch example, use domain breaks and derivative signs; this workshop does not replace that lesson or the printed-panel problem above.

A complete investigation

Investigate $f(x)=x^2/(x-1)$. Find its domain, intercepts, asymptotes, variation, extrema and concavity. Sketch separate branches and justify whether any inflection point exists. Try an initial sketch before reading the explanation.

Read the complete investigation

1. Record the domain and intercepts

The denominator excludes $x=1$: $D=\mathbb R\setminus\{1\}$. The only zero of the numerator is $x=0$, which is allowed, so the x- and y-intercepts coincide at $(0,0)$. Do not erase the excluded input when making sign tables.

2. Separate the polynomial part

Polynomial division gives $$f(x)=x+1+\frac1{x-1}.$$ As $x\to1^-$, the last term tends to $-\infty$; as $x\to1^+$, it tends to $+\infty$. Thus $x=1$ is a vertical asymptote. Since $f(x)-(x+1)\to0$ at both infinities, the slant asymptote is $y=x+1$, not a horizontal line. The graph stays below this line for $x<1$ and above it for $x>1$, because the difference is $1/(x-1)$. It never crosses the line in this example.

3. Use signs, not just stationary inputs

$$f'(x)=1-\frac1{(x-1)^2}=\frac{x(x-2)}{(x-1)^2}.$$ The denominator is positive wherever defined. Split at both zeros, 0 and 2, and at the excluded input 1.

• $x<0$: $f'>0$, increasing toward $(0,0)$.
• $0<x<1$: $f'<0$, decreasing toward $-\infty$.
• $1<x<2$: $f'<0$, decreasing from $+\infty$.
• $x>2$: $f'>0$, increasing away from $(2,4)$.

The sign change + to − proves a local maximum at $(0,0)$; − to + proves a local minimum at $(2,4)$. Neither is global on the full domain: values are unbounded above and below. Do not describe the curve as decreasing “from 0 to 2” without the break at 1.

4. Keep concavity separate from increasing/decreasing

$$f''(x)=\frac2{(x-1)^3}.$$ It is negative on $(-\infty,1)$, so the left branch is concave down; it is positive on $(1,\infty)$, so the right branch is concave up. The change happens across a missing point, so there is no inflection point. A point of inflection must lie on the graph.

5. Assemble and check the sketch

Text equivalent: the left branch approaches the slant line from below as x tends to negative infinity, rises to (0,0), then falls to negative infinity at 1 from the left. The right branch starts at positive infinity at 1 from the right, falls to (2,4), then rises while approaching the slant line from above. The left branch is concave down throughout; the right is concave up throughout.

Check $f(-1)=-1/2$, $f(1/2)=-1/2$, $f(3)=9/2$. Sample points help catch plotting errors, but the derivative signs justify the complete branch behaviour.

Three transfer problems

Keep the answers closed. For each conclusion, write the evidence: domain, a sign change, a limit or a slope on the derivative graph.

1. Build a fresh sign table

Let $g(x)=(x+1)^2/(x-2)$. Without graphing software, find the domain, slant asymptote and all increasing/decreasing intervals. Locate and classify the local extrema; explain whether they are global. You may use $g'(x)=(x+1)(x-5)/(x-2)^2$, but verify it by rewriting g first.

2. Read the derivative’s graph, not the function’s height

A differentiable function F on $(-3,5)$ has the continuous derivative shown below: straight segments join $(-3,2)$, $(1,-2)$, and $(5,2)$. In particular, the derivative crosses zero at −1 and 3. Given $F(1)=2$, find the increasing/decreasing intervals, local extrema by input, concavity intervals and inflection point. No integration is needed; the function value at 1 is supplied. Do not mistake a vertex of F′ for an extremum of F.

3. Diagnose an inflection-point claim

A student says: “If the second derivative is zero or changes sign at an input, that input is an inflection point.” Test the claim at 0 for $p(x)=x^4$, $q(x)=1/x$, and $r(x)=x^3$. State which graph, if any, has an inflection point there, and give a reason for each decision.

Choose your next attempt: if an excluded input disappeared from your table, revisit domains and limits; if you confused increasing with concave up, repeat problem 2 with separate statements about F′ and F″. For tutoring, bring your section’s scope, your sketch and the first step you cannot justify. Keep using the course’s existing tutoring request rather than assuming an exam topic or a confirmed appointment.