Test 3 Review Material

Below are resources to help you practice and prepare for the test 3 in this course.

Test 3 will cover the material covered in Chapter 5 of this textbook:

understanding what sign charts tell us about the function
ability to identify local and global extrema, points of inflection and curvature
ability to use a sign chart to find extrema
ablity to plot a function based on information from the first and second derivative sign charts

Test Instructions

Below are the general instructions for all tests

You will get a chance to retake this. Your highest score counts toward your final grade.
Follow the guidance for each part and show all work for full credit.
Non-graphical calculators are allowed.
One page (two sides) of handwritten (or font size 8) notes are allowed.
The equation sheet from the textbook is included as part of the test pack (it doesn’t count as note pages)

Practice Short Answer

If \(f'(x) > 0\) on an interval, what does this tell you about how the environmental variable is behaving? Give an example of a real system that might show this pattern.

Suggested Answer

The variable is increasing on that interval. Examples include lake temperature rising through the morning, snowpack depth increasing during a storm, or early-stage population growth.

A population model has a point where \(P'(t) = 0\) but \(P''(t) < 0\). Explain what is happening biologically at that moment.

Suggested Answer

The population has reached a peak. Growth has slowed to zero, and the negative second derivative shows decline is beginning due to limits like resources or space.

Why is an inflection point important in environmental systems? Describe a situation where identifying the moment when concavity changes could be useful.

Suggested Answer

An inflection point marks when the rate of change switches from accelerating to decelerating. Useful examples include algae transitioning from rapid bloom to nutrient-limited slowdown or a streamflow peak transitioning to recession.

If a function has a horizontal asymptote as \(x \to +\infty\), what does that usually represent in ecological or physical models?

Suggested Answer

A long-term limiting value. Examples: carrying capacity, maximum nutrient uptake, equilibrium storage or temperature.

A function has a vertical asymptote at \(x = 3\). What does this mean, and what environmental boundary might behave this way?

Suggested Answer

The function becomes undefined at \(x=3\) and approaches ±∞. Environmental analogies include thresholds like pore pressure reaching critical limits or denominators representing zero volume remaining.

You are given only a derivative graph \(f'(x)\). How can you determine where the original function is increasing or decreasing, has maxima/minima, or is concave up/down?

Suggested Answer

Increasing where \(f'>0\), decreasing where \(f'<0\). A max occurs where \(f'\) changes +→– and a min where –→+. Concavity depends on whether \(f'\) is rising (concave up) or falling (concave down).

The second derivative of a function is positive for all \(x\). What does this tell you about the shape of the curve?

Suggested Answer

The function is concave up everywhere. This could represent accelerating recovery or compounding growth after disturbance.

A logistic-style function changes concavity at exactly one value of \(x\). What does that point represent in system dynamics?

Suggested Answer

It is the inflection point—the moment of maximum growth rate, where growth switches from accelerating to slowing.

If \(f'(x)\) is always positive but decreases toward zero as \(x \to \infty\), how would the graph look? What real process behaves this way?

Suggested Answer

The function increases but levels off toward a horizontal asymptote. Examples include nutrient uptake saturation, long-term warming trends, or logistic population limits.

An environmental dataset increases, then decreases, then increases again. What must the sign of the derivative look like?

Suggested Answer

The derivative must be +, then –, then +. Causes could include seasonal cycles, predator–prey oscillations, or storm hydrograph rise–fall–rise sequences.

If two functions both satisfy \(f'(a)=0\), must they have the same type of extremum at \(x=a\)? Explain.

Suggested Answer

No. A horizontal tangent could be a max, min, saddle, or flat point. The behavior depends on sign changes in \(f'\) or on \(f''(a)\).

A function’s rate of change increases rapidly, then levels off. What does this say about \(f'\) and \(f''\)?

Suggested Answer

\(f'>0\) throughout. \(f''>0\) during the acceleration phase, then \(f''<0\) as growth slows. This is typical of logistic or saturation processes.

In an environmental time series, what does it mean if \(f''(t) < 0\) while \(f'(t) > 0\)?

Suggested Answer

The variable is increasing but at a slowing rate. Examples: warming lake approaching equilibrium, population nearing carrying capacity.

Suppose a function has an inflection point at \(x=5\). What does this tell you about system behavior on each side?

Suggested Answer

Concavity switches sign at 5. The system moves from accelerating to slowing change (or vice versa), marking a transition in dynamics.

Why is a vertical asymptote different from a point where the derivative is undefined?

Suggested Answer

A vertical asymptote means the function itself blows up to ±∞. A derivative can be undefined even when the function is finite (corners, cusps). These reflect different structural vs. behavioral limitations.

A streamflow model reaches a maximum at \(t=7\) hours. What does this say about the hydrologic process?

Suggested Answer

Before 7 hr, discharge increases; after, it decreases. This marks peak flow, when routing and inputs balance before recession dominates.

A derivative graph shows a long flat region where \(f'(x) \approx 0\). What might this indicate?

Suggested Answer

The system is experiencing very little change. Examples: midday temperature plateau, stable soil moisture, population near carrying capacity.

Why is the quadratic formula useful when analyzing derivatives?

Suggested Answer

Many derivatives simplify to quadratics. Finding extrema or inflection points requires solving \(f'=0\) or \(f''=0\), and the quadratic formula works even when factoring doesn’t.

If a function always has a small but positive derivative, how would you describe the shape?

Suggested Answer

It increases gradually with no peaks or sharp transitions. Examples include slow climate warming, gradual biomass growth, or long-term CO₂ trends.

A model has a horizontal asymptote at \(y=120\). What does this say about long-term system behavior?

Suggested Answer

The system approaches a stable upper limit, such as a carrying capacity, saturation level, or maximum equilibrium value.

A function has
\[ \lim_{x\to\infty} f(x)=8. \]
What does this horizontal asymptote tell you about the long-term behavior of the system being modeled?

Suggested Answer

The system approaches and stabilizes around 8 as the input variable becomes very large. This represents a long-term equilibrium such as a carrying capacity, saturation point, or thermal equilibrium.

Given
\[ f(x)=\frac{5x}{x-4}, \]
identify the vertical asymptote and describe the behavior of the function as \(x\) approaches that asymptote from the left and from the right.

Suggested Answer

The vertical asymptote is at \(x=4\).
As \(x\to 4^-\), the denominator is negative and very small, so \(f(x)\to -\infty\).
As \(x\to 4^+\), the denominator is positive and very small, so \(f(x)\to +\infty\).

Explain the difference between approaching a horizontal asymptote from above vs. approaching it from below.
What does each imply about the sign of the derivative for large \(x\)?

Suggested Answer

Approaching from above means the function is decreasing toward the asymptote, so \(f'(x) < 0\) for large \(x\).
Approaching from below means the function is increasing toward the asymptote, so \(f'(x) > 0\) for large \(x\).

A function has two vertical asymptotes at \(x=-1\) and \(x=3\).
What does this imply about the domain and structure of the graph?

Suggested Answer

The domain excludes \(x=-1\) and \(x=3\).
The graph is broken into three disconnected pieces, each with potentially different behavior (increasing/decreasing, concavity, extrema), since the function cannot cross those asymptotes.

A rational function approaches the horizontal asymptote \(y=0\) as \(x\to\pm\infty\).
Give a real-world environmental process that might behave this way.

Suggested Answer

Examples include pollutant concentration decaying toward zero, cooling curves approaching ambient temperature, radioactive decay processes, or stream velocity decreasing to near zero as flood waters recede.

Practice Problems

Problem 1 — Clean Cubic Example

Consider the function
\[ f(x) = x^3 - 3x. \]

Use derivatives and sign charts to analyze its behavior.

Check 1st derivative

\[ f'(x) = 3(x-1)(x+1). \]

Check critical points

\[ f'(x)=0 \quad\Rightarrow\quad x=-1,\;1. \]

Check 2nd derivative

\[ f''(x)=6x. \]

Check inflection point

\[ f''(x)=0 \quad\Rightarrow\quad x=0. \]

Full solution

Start with
\[ f(x)=x^3 - 3x. \]

1. First derivative (with factoring)
\[ f'(x)=3x^2 - 3. \]

Factor the derivative: \[ 3x^2 - 3 = 3(x^2 - 1) = 3(x-1)(x+1). \]

Critical points solve
\[ 3(x-1)(x+1)=0 \quad\Rightarrow\quad x=-1,\;1. \]

Sign of the derivative: - For \(x<-1\): both factors negative → \(f'(x)>0\)
- For \(-1<x<1\): one positive, one negative → \(f'(x)<0\)
- For \(x>1\): both positive → \(f'(x)>0\)

So the function is: - Increasing on \((-\infty,-1)\)
- Decreasing on \((-1,1)\)
- Increasing on \((1,\infty)\)

Thus: - Local maximum at \(x=-1\)
- Local minimum at \(x=1\)

2. Second derivative (with factoring)
\[ f''(x)=6x. \]

Solve \(f''(x)=0\): \[ 6x=0 \quad\Rightarrow\quad x=0. \]

Sign of second derivative: - \(f''(x)<0\) for \(x<0\) → concave down
- \(f''(x)>0\) for \(x>0\) → concave up

Thus \(x=0\) is an inflection point.

Putting together the behavior:

Increasing → decreasing → increasing
Concave down for \(x<0\)
Concave up for \(x>0\)
Local max at \(x=-1\)
Local min at \(x=1\)
Inflection at \(x=0\)

Problem 2 — Symmetric Quartic

Consider the function
\[ f(x) = x^4 - 4x^2. \]

Use derivatives and sign charts to analyze its behavior.

Check 1st derivative

\[ f'(x) = 4x(x^2 - 2). \]

Check critical points

\[ x = 0,\quad x = \pm\sqrt{2}. \]

Check 2nd derivative

\[ f''(x) = 4(3x^2 - 2). \]

Check inflection point

\[ x = \pm\sqrt{\tfrac{2}{3}}. \]

Full solution

Start with
\[ f(x)=x^4 - 4x^2. \]

1. First derivative (with factoring)
\[ f'(x)=4x^3 - 8x = 4x(x^2 - 2). \]

Critical points solve
\[ 4x(x^2 - 2) = 0 \quad\Rightarrow\quad x = 0,\; \pm\sqrt{2}. \]

Sign of derivative: - For \(x < -\sqrt{2}\): \(f'(x) < 0\)
- For \(-\sqrt{2} < x < 0\): \(f'(x) > 0\)
- For \(0 < x < \sqrt{2}\): \(f'(x) < 0\)
- For \(x > \sqrt{2}\): \(f'(x) > 0\)

Thus: - Local max at \(x = 0\)
- Local mins at \(x = \pm\sqrt{2}\)

2. Second derivative (with factoring)
\[ f''(x)=12x^2 - 8 = 4(3x^2 - 2). \]

Solve
\[ 3x^2 - 2 = 0 \quad\Rightarrow\quad x = \pm\sqrt{\tfrac{2}{3}}. \]

Concavity: - Down for \(|x| < \sqrt{2/3}\)
- Up for \(|x| > \sqrt{2/3}\)

Problem 3 — Rational Function With Simple Roots

Consider the function
\[ f(x) = \frac{x}{x^2 + 1}. \]

Use derivatives and sign charts to analyze its behavior.

Check 1st derivative

\[ f'(x) = \frac{1 - x^2}{(x^2+1)^2}. \]

Check critical points

\[ x = \pm 1. \]

Check 2nd derivative

\[ f''(x) = \frac{-2x(x^2+3)}{(x^2+1)^3}. \]

Check inflection point

\[ x = 0. \]

Full solution

Start with
\[ f(x)=\frac{x}{x^2+1}. \]

1. First derivative (with factoring)

Apply the quotient rule:

\[ f'(x)=\frac{(1)(x^2+1) - x(2x)}{(x^2+1)^2} = \frac{x^2+1 - 2x^2}{(x^2+1)^2} = \frac{1 - x^2}{(x^2+1)^2}. \]

Critical points solve
\[ 1 - x^2 = 0 \quad\Rightarrow\quad x = \pm 1. \]

Derivative sign:

\(x < -1\): \(1 - x^2 < 0\) → decreasing
\(-1 < x < 1\): \(1 - x^2 > 0\) → increasing
\(x > 1\): \(1 - x^2 < 0\) → decreasing

Thus: - Local min at \(x=-1\)
- Local max at \(x=1\)

2. Second derivative (with factoring)

\[ f''(x) = \frac{-2x(x^2+3)}{(x^2+1)^3}. \]

Set numerator to zero:

\[ -2x(x^2+3)=0. \]

Since \(x^2+3 > 0\),
\[ x = 0 \] is the only inflection point.

Concavity:

\(x < 0\): \(f''(x) > 0\) → concave up
\(x > 0\): \(f''(x) < 0\) → concave down

Problem 4 — Exponential Product With Clean Factors

Consider the function
\[ f(x) = x^2 e^{-x}. \]

Use derivatives and sign charts to analyze its behavior.

Check 1st derivative

\[ f'(x) = e^{-x}x(2 - x). \]

Check critical points

\[ x = 0,\quad x = 2. \]

Check 2nd derivative

\[ f''(x) = e^{-x}(x^2 - 4x + 2). \]

Check inflection points

\[ x = 2 \pm \sqrt{2}. \]

Full solution

Start with
\[ f(x)=x^2 e^{-x}. \]

1. First derivative (with factoring)

Use the product rule:
\[ f'(x) = (2x)e^{-x} + x^2(-e^{-x}) = e^{-x}(2x - x^2) = e^{-x}x(2 - x). \]

Critical points come from
\[ e^{-x}x(2-x)=0. \]

Since \(e^{-x} \neq 0\), solutions are
\[ x=0,\quad x=2. \]

Sign of \(f'(x)\):

For \(x<0\): \(x<0\), \(2-x>0\) → negative → decreasing
For \(0<x<2\): positive × positive → positive → increasing
For \(x>2\): positive × negative → negative → decreasing

Thus: - Local minimum at \(x=0\)
- Local maximum at \(x=2\)

2. Second derivative (with factoring)

Differentiate
\[ f'(x)=e^{-x}(2x - x^2). \]

Using product rule again: \[ f''(x)=e^{-x}(-1)(2x - x^2) + e^{-x}(2 - 2x) = e^{-x}(x^2 - 4x + 2). \]

Solve
\[ x^2 - 4x + 2 = 0. \]

Roots: \[ x = 2 \pm \sqrt{2}. \]

Concavity:

Up on \((-\infty,\, 2-\sqrt{2})\)
Down on \((2-\sqrt{2},\, 2+\sqrt{2})\)
Up on \((2+\sqrt{2},\,\infty)\)

Problem 5 — Logarithm With Symmetry

Consider the function
\[ f(x) = \ln(x^2 + 1). \]

Use derivatives and sign charts to analyze its behavior.

Check 1st derivative

\[ f'(x)=\frac{2x}{x^2+1}. \]

Check critical points

\[ f'(x)=0 \quad\Rightarrow\quad x=0. \]

Check 2nd derivative

\[ f''(x)=\frac{2(1 - x^2)}{(x^2+1)^2}. \]

Check inflection points

\[ x = \pm 1. \]

Full solution

Start with
\[ f(x)=\ln(x^2+1). \]

1. First derivative (with factoring)

Use chain rule:
\[ f'(x)=\frac{1}{x^2+1}\cdot 2x = \frac{2x}{x^2+1}. \]

Critical point from
\[ \frac{2x}{x^2+1}=0 \quad\Rightarrow\quad x=0. \]

Sign of \(f'(x)\): - For \(x<0\): numerator negative → decreasing
- For \(x>0\): numerator positive → increasing

Thus the function is: - Decreasing on \((-\infty,0)\)
- Increasing on \((0,\infty)\)

So \(x=0\) is a local minimum.

2. Second derivative (with factoring)

Differentiate
\[ f'(x)=\frac{2x}{x^2+1}. \]

Using quotient rule: \[ f''(x)=\frac{2(x^2+1) - 2x(2x)}{(x^2+1)^2} =\frac{2 - 2x^2}{(x^2+1)^2} =\frac{2(1-x^2)}{(x^2+1)^2}. \]

Solve
\[ 1 - x^2 = 0 \quad\Rightarrow\quad x=\pm 1. \]

Concavity: - For \(|x|<1\): \(1-x^2>0\) → concave up
- For \(|x|>1\): \(1-x^2<0\) → concave down

Thus \(x=\pm 1\) are inflection points.

Problem 6 — Rational Function With a Vertical Asymptote

Consider the function
\[ f(x) = \frac{x^2 - 1}{x - 2}. \]

Use derivatives, sign charts, and asymptotes to analyze its behavior.

Check 1st derivative

\[ f'(x)=\frac{x^2 - 4x + 1}{(x-2)^2}. \]

Check critical points

\[ x = 2 \pm \sqrt{3}. \]

Check 2nd derivative

\[ f''(x)=\frac{6}{(x-2)^3}. \]

Check inflection points

None — concavity switches only across the asymptote.

Full solution

We analyze
\[ f(x)=\frac{x^2 - 1}{x - 2}. \]

1. Domain and asymptote

The denominator is zero at
\[ x = 2, \]
so the domain is all real numbers except \(x=2\).
Thus there is a vertical asymptote at \(x=2\).

2. First derivative (with factoring)

Quotient rule gives
\[ f'(x)=\frac{x^2 - 4x + 1}{(x-2)^2}. \]

Critical points come from the numerator:

\[ x^2 - 4x + 1=0 \quad\Rightarrow\quad x = 2 \pm \sqrt{3}. \]

The derivative is undefined at the vertical asymptote \(x=2\).

3. Second derivative (with factoring)

\[ f''(x)=\frac{6}{(x-2)^3}. \]

Sign analysis:

For \(x<2\): denominator \(<0\) → \(f''(x)<0\) → concave down
For \(x>2\): denominator \(>0\) → \(f''(x)>0\) → concave up

No inflection points occur within a domain interval — concavity only changes across the asymptote.

4. Behavior near the vertical asymptote

To determine the direction of blow-up, compute limits from both sides.

Left-hand limit (approach from below)

As \(x\to 2^-\):

Numerator \(x^2 - 1 \to 3\) (positive)
Denominator \(x - 2 \to 0^-\) (negative)

Thus
\[ \lim_{x\to 2^-} f(x) = \frac{3}{0^-} = -\infty. \]

Right-hand limit (approach from above)

As \(x\to 2^+\):

Numerator still \(x^2 - 1 \to 3\)
Denominator \(x - 2 \to 0^+\) (positive)

Thus
\[ \lim_{x\to 2^+} f(x) = \frac{3}{0^+} = +\infty. \]

5. Summary of behavior

Vertical asymptote at \(x=2\)
- \(f(x)\to -\infty\) as \(x\to 2^-\)
- \(f(x)\to +\infty\) as \(x\to 2^+\)
Critical points at
\[ x=2-\sqrt{3},\quad x=2+\sqrt{3}. \]
Concavity:
- Down on \((-\infty, 2)\)
- Up on \((2, \infty)\)

Problem 7 — Rational Function With Two Vertical Asymptotes

Consider the function
\[ f(x) = \frac{1}{x(x-2)}. \]

Use derivatives, sign charts, and asymptotes to analyze its behavior.

Check 1st derivative

\[ f'(x)= -\frac{2(x-1)}{x^2(x-2)^2}. \]

Check critical points

\[ x = 1. \]

Check 2nd derivative

\[ f''(x)=\frac{2(3x^2 - 6x + 4)}{x^3 (x-2)^3}. \]

Check inflection points

Solve
\[ 3x^2 - 6x + 4 = 0 \]

No REAL roots

Full solution

We analyze
\[ f(x)=\frac{1}{x(x-2)}. \]

The denominator is zero at
\[ x = 0,\qquad x = 2, \] so these produce vertical asymptotes and the domain is
\[ (-\infty,0)\;\cup\;(0,2)\;\cup\;(2,\infty). \]

To compute the first derivative, rewrite the function as
\[ f(x)=[x(x-2)]^{-1}. \] Differentiating gives
\[ f'(x)=-(x(x-2))^{-2}(2x-2) =-\frac{2(x-1)}{x^2(x-2)^2}. \] Critical points come from the numerator: \[ -2(x-1)=0 \quad\Rightarrow\quad x=1. \] The derivative is undefined at the vertical asymptotes.

The second derivative is
\[ f''(x)=\frac{2(3x^2 - 6x + 4)}{x^3 (x-2)^3}. \] Second-derivative critical points solve
\[ 3x^2 - 6x + 4=0. \] Using the quadratic formula, \[ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} =\frac{6\pm\sqrt{36-48}}{6} =\frac{6\pm\sqrt{-12}}{6}. \] Because the discriminant is negative, the numerator never equals zero, so there are no real inflection points from the second derivative. Concavity changes only when crossing the asymptotes since the denominator changes sign there.

To understand the behavior near \(x=0\):

As \(x\to 0^-:\)
denominator \(0^-\cdot(-2)=0^+\), so
\[ \lim_{x\to 0^-} f(x)=+\infty. \]

As \(x\to 0^+:\)
denominator \(0^+\cdot(-2)=0^-\), so
\[ \lim_{x\to 0^+} f(x)=-\infty. \]

At \(x=2\):

As \(x\to 2^-:\)
\((2^-)(0^-)=0^{+}\), so
\[ \lim_{x\to 2^-} f(x)=+\infty. \]

As \(x\to 2^+:\)
\((2^+)(0^+)=0^{+}\), so
\[ \lim_{x\to 2^+} f(x)=+\infty. \]

For the horizontal asymptotes, note that
\[ f(x)=\frac1{x(x-2)}\sim \frac1{x^2} \] as \(x\to\pm\infty\). Thus
\[ \lim_{x\to\pm\infty} f(x)=0, \] so the horizontal asymptote is
\[ y=0. \]

In summary:

Vertical asymptotes at \(x=0\) and \(x=2\).
Horizontal asymptote \(y=0\).
One critical point from the first derivative at \(x=1.\)
No real inflection points because the quadratic in the numerator of \(f''\) has no real roots.
Concavity changes only across asymptotes, due to the denominator’s sign.

Problem 8 — Logistic-Style Function With Two Horizontal Asymptotes

Consider the function
\[ f(x)=\frac{80}{1+3e^{-0.5x}}. \]

Use derivatives and asymptotes to analyze its behavior.

Check 1st derivative

\[ f'(x)=\frac{120e^{-0.5x}}{(1+3e^{-0.5x})^2}. \]

Check critical points

The derivative is always positive.
\[ \text{No critical points (always increasing).} \]

Check 2nd derivative

\[ f''(x)=\frac{60e^{-0.5x}(3e^{-0.5x}-1)}{(1+3e^{-0.5x})^3}. \]

Check inflection point

Solve
\[ 3e^{-0.5x}-1=0. \]

\[ e^{-0.5x}=\tfrac13. \]

\[ x = 2\ln(3). \]

Check horizontal asymptotes

\[ \lim_{x\to-\infty} f(x)=0, \qquad \lim_{x\to+\infty} f(x)=80. \]

Horizontal asymptotes: \[ y=0, \qquad y=80. \]

Full solution

We analyze
\[ f(x)=\frac{80}{1+3e^{-0.5x}}. \]

Rewrite the function as
\[ f(x)=80(1+3e^{-0.5x})^{-1}. \]

Differentiate:

\[ f'(x)=80(-1)(1+3e^{-0.5x})^{-2}\cdot(3e^{-0.5x}\cdot -0.5) =\frac{120e^{-0.5x}}{(1+3e^{-0.5x})^2}. \]

Since all factors are positive,
\[ f'(x) > 0 \] for every real \(x\).
Thus the function is always increasing and has no local maxima or minima.

Differentiate again:

\[ f''(x)=\frac{60e^{-0.5x}(3e^{-0.5x}-1)}{(1+3e^{-0.5x})^3}. \]

Set the numerator equal to zero:

\[ 3e^{-0.5x}-1=0 \]

\[ e^{-0.5x}=\tfrac13 \]

\[ x = 2\ln(3). \]

For \(e^{-0.5x} > \frac13\), the graph is concave up;
for \(e^{-0.5x} < \frac13\), the graph is concave down.

As \(x\to -\infty\):

\(e^{-0.5x}\to\infty\), denominator → ∞
\[ f(x)\to 0. \]

As \(x\to +\infty\):

\(e^{-0.5x}\to 0\), denominator → 1
\[ f(x)\to 80. \]

Thus the function is an increasing logistic curve with a lower horizontal asymptote at \(y=0\) and an upper horizontal asymptote at \(y=80\), switching concavity at the point \(x = 2\ln 3\).

Challenge 1 — Rational Function

Consider the function
\[ f(x) = \frac{x^2 - 1}{x^2 + 1}. \]

Analyze the first derivative, second derivative, critical points, and inflection points.

Check 1st derivative

\[ f'(x) = \frac{4x}{(x^2+1)^2}. \]

Check critical points

Critical points solve \(f'(x)=0\):

\[ 4x = 0 \quad\Rightarrow\quad x = 0. \]

Check 2nd derivative

\[ f''(x) = 4\frac{1 - 3x^2}{(x^2+1)^3}. \]

Check inflection points

Solve \(1 - 3x^2 = 0\):

\[ x = \pm \frac{1}{\sqrt{3}}. \]

Full solution

We analyze
\[ f(x)=\frac{x^2 - 1}{x - 2}. \]

1. Domain and asymptote

\[ x - 2 = 0 \quad\Rightarrow\quad x = 2. \]

Thus the domain is
\[ (-\infty,2)\cup(2,\infty), \] with a vertical asymptote at \(x=2\).

2. First derivative (with factoring)

Using the quotient rule:

\[ f'(x)=\frac{(2x)(x-2) - (x^2 - 1)(1)}{(x-2)^2}. \]

Expand the numerator:

\[ 2x(x-2)=2x^2 - 4x, \]

Subtract the second term:

\[ 2x^2 - 4x - (x^2 - 1) = x^2 - 4x + 1. \]

So the derivative is
\[ f'(x)=\frac{x^2 - 4x + 1}{(x-2)^2}. \]

How the critical points were found

Critical points occur when the derivative equals zero and is defined.

The denominator \((x-2)^2\) is never zero except at \(x=2\), where the function is undefined.
So we only solve the numerator:

\[ x^2 - 4x + 1 = 0. \]

Apply the quadratic formula:

\[ x=\frac{4 \pm \sqrt{(-4)^2 - 4(1)(1)}}{2} =\frac{4 \pm \sqrt{16 - 4}}{2} =\frac{4 \pm \sqrt{12}}{2} =\frac{4 \pm 2\sqrt{3}}{2} =2 \pm \sqrt{3}. \]

Thus the critical points are:

\[ x=2-\sqrt{3},\qquad x=2+\sqrt{3}. \]

Because the derivative does not exist at \(x=2\), that point is not a critical point.

3. Second derivative

We start with
\[ f'(x)=\frac{x^2 - 4x + 1}{(x-2)^2}. \]

Using the quotient rule again:

\[ f''(x)=\frac{6}{(x-2)^3}. \]

(Steps omitted here since they were checked previously.)

Sign of \(f''(x)\):

For \(x<2\): denominator negative → concave down
For \(x>2\): denominator positive → concave up

Concavity switches across the vertical asymptote, not within the domain, so there are no inflection points.

4. Limits at the vertical asymptote

Left-hand limit: \[ \lim_{x\to 2^-} f(x) =\frac{x^2 - 1}{x - 2} =\frac{3}{0^-} =-\infty. \]

Right-hand limit: \[ \lim_{x\to 2^+} f(x) =\frac{3}{0^+} =+\infty. \]

Thus:

\(f(x)\to -\infty\) from the left of the asymptote
\(f(x)\to +\infty\) from the right of the asymptote

5. Summary

Vertical asymptote at \(x=2\)
Critical points at \(x = 2\pm\sqrt{3}\)
Concave down on \((-\infty,2)\), concave up on \((2,\infty)\)
Behavior flips dramatically across the asymptote

Practice Test — Sign Charts & Applications

This practice set mirrors the structure of your problems on the test:

Section A — Extrema from a factored first derivative
Section B — Sign charts + sketching using \(f'\) and \(f''\)
Section C — Asymptotes, extrema, concavity and sketching

Work through each table as if each row were its own problem.

Section A — Extrema from a Factored First Derivative (Practice)

Use the provided factored derivative to:

Identify critical points
Create a first-derivative sign chart
Classify each critical point (local max/min/neither)

Function	Factored Derivative \(f'(x)\)
\(f_1(x)\)	\(f_1'(x)=e^{x}(x-1)(x+4)\)
\(f_2(x)\)	\(f_2'(x)=-(x+2)(x-5)(x-1)\)
\(f_3(x)\)	\(f_3'(x)=\dfrac{(x+3)(x-2)}{(x^2+1)}\)

Solution Key Section A (click to expand)

For each function we:

Identify critical points
Build a sign chart
Classify each point

✔ Function \(f_1(x)\)
\[ f_1'(x)=e^{x}(x-1)(x+4) \]

Critical points:
\[ x=-4,\qquad x=1 \]

Sign chart:

Interval	Sign of \(f_1'(x)\)	Behavior
\(x<-4\)	\((-)(-)=+\) → \(+\)	Increasing
\(-4<x<1\)	\((+)(-)= -\)	Decreasing
\(x>1\)	\((+)(+)=+\)	Increasing

Classification:
- At \(x=-4\): increasing → decreasing → local max
- At \(x=1\): decreasing → increasing → local min

✔ Function \(f_2(x)\)
\[ f_2'(x)=-(x+2)(x-5)(x-1) \]

Critical points:
\[ x=-2,\quad x=1,\quad x=5 \]

Sign chart (including the leading negative):

Interval	Sign of product	After negative sign	Behavior
\(x<-2\)	\((-)(-)(-)= -\)	\(+\)	Increasing
\(-2<x<1\)	\((+)(-)(-)=+\)	\(-\)	Decreasing
\(1<x<5\)	\((+)(-)(+)= -\)	\(+\)	Increasing
\(x>5\)	\((+)(+)(+)=+\)	\(-\)	Decreasing

Classification:

\(x=-2\): inc → dec → local max
\(x=1\): dec → inc → local min
\(x=5\): inc → dec → local max

✔ Function \(f_3(x)\)
\[ f_3'(x)=\frac{(x+3)(x-2)}{x^2+1} \]

Denominator always positive → sign from numerator only.

Critical points: \[ x=-3,\qquad x=2 \]

Sign chart:

Interval	Sign of numerator	Behavior
\(x<-3\)	\((-)(-)=+\)	Increasing
\(-3<x<2\)	\((+)(-)= -\)	Decreasing
\(x>2\)	\((+)(+)=+\)	Increasing

Classification:

\(x=-3\): local max
\(x=2\): local min

Section B — Sign Charts & Sketching from \(f'\) and \(f''\)

For each function, use the given derivatives to:

Find and classify critical points
Determine intervals of increasing/decreasing
Determine concavity
Sketch a possible graph of the function

Function	First Derivative \(f'(x)\)	Second Derivative \(f''(x)\)
\(g_1(x)\)	\(g_1'(x)= (x-2)(x+1)\)	\(g_1''(x)=2x-3\)
\(g_2(x)\)	\(g_2'(x)= x(x+5)\)	\(g_2''(x)=2x+5\)
\(g_3(x)\)	\(g_3'(x)= (x-4)(x-1)(x+2)\)	\(g_3''(x)=3x^2-6\)

Solution Key Section B (click to expand)

✔ Function \(g_1(x)\)
\[ g_1'(x)= (x-2)(x+1),\qquad g_1''(x)=2x-3 \]

Critical points:
\[ x=-1,\quad x=2 \]

First derivative sign chart:

\(x<-1\): inc
\(-1<x<2\): dec
\(x>2\): inc

→ Local max at \(-1\)
→ Local min at \(2\)

Second derivative:

\[ g_1''(x)=0 \Rightarrow x=\frac{3}{2}=1.5 \]

Concavity:

\(x<1.5\): \(g_1''<0\) → concave down
\(x>1.5\): concave up

Sketch structure:

Peak at \(-1\)
Valley at \(2\)
Concave down until \(1.5\), concave up afterward

✔ Function \(g_2(x)\)
\[ g_2'(x)= x(x+5),\qquad g_2''(x)=2x+5 \]

Critical points:
\[ x=0,\quad x=-5 \]

Sign chart:

\(x<-5\): inc
\(-5<x<0\): dec
\(x>0\): inc

→ Local max at \(x=-5\)
→ Local min at \(x=0\)

Second derivative:

\[ g_2''(x)=0 \Rightarrow x=-\frac52=-2.5 \]

Concavity:

\(x<-2.5\): concave down
\(x>-2.5\): concave up

Sketch:

A classic peak → valley → rising curve.

✔ Function \(g_3(x)\)
\[ g_3'(x)= (x-4)(x-1)(x+2),\qquad g_3''(x)=3x^2-6 \]

Critical points:
\[ x=-2,\quad x=1,\quad x=4 \]

Sign chart:

\(x<-2\): decreasing
\(-2<x<1\): increasing
\(1<x<4\): decreasing
\(x>4\): increasing

→ Local min at \(-2\)
→ Local max at \(1\)
→ Local min at \(4\)

Second derivative:

\[ g_3''(x)=0 \Rightarrow 3x^2-6=0 \Rightarrow x=\pm \sqrt{2} \]

Concavity:

\(x<-\sqrt2\): up
\(-\sqrt2<x<\sqrt2\): down
\(x>\sqrt2\): up

Sketch:

Two valleys, one peak, with inflection points at \(\pm \sqrt{2}\).

Section C — Asymptotes, Extrema, and Concavity

For each function:

Identify vertical asymptotes
Identify horizontal asymptotes using limits
Use the provided first derivative to make a first-derivative sign chart
Use the second derivative to determine concavity
Describe end behavior
Sketch the qualitative shape

Table C1 — Functions & First Derivatives

Function	First Derivative \(h'(x)\)
\(h_1(x)=\dfrac{2e^{-x}}{x-3}+1\)	\(h_1'(x)=\dfrac{-2e^{-x}(x-3)-2e^{-x}}{(x-3)^2}\)
\(h_2(x)=\dfrac{5}{x^{2}-4}+2\)	\(h_2'(x)=\dfrac{-10x}{(x^{2}-4)^{2}}\)
\(h_3(x)=\dfrac{e^{x}}{x+1}-4\)	\(h_3'(x)=\dfrac{e^{x}(x+1)-e^{x}}{(x+1)^2}=\dfrac{e^{x}x}{(x+1)^2}\)

Table C2 — Second Derivatives & Asymptotes

Second Derivative \(h''(x)\)	Asymptotes
\(h_1''(x)=\dfrac{2e^{-x}(x^{2}-6x+11)}{(x-3)^{3}}\)	VA: \(x=3\) HA: \(y=1\)
\(h_2''(x)=\dfrac{30x^{2}+40}{(x^{2}-4)^{3}}\)	VA: \(x=\pm 2\) HA: \(y=2\)
\(h_3''(x)=\dfrac{e^{x}(x^{2}+1)}{(x+1)^{3}}\)	VA: \(x=-1\) HA: \(y=-4\)

Solution Key Section C (click to expand)

We use both C1 and C2 tables.

✔ Solution: Function \(h_1(x)=\dfrac{2e^{-x}}{x-3}+1\)

1. Vertical Asymptote

The denominator is zero at \(x=3\):

\[ \text{Vertical asymptote: } x=3 \]

2. Horizontal Asymptote

As \(x\to\pm\infty\), the exponential term dies out:

\[ \lim_{x\to\infty} h_1(x)=1, \qquad \lim_{x\to -\infty} h_1(x)=1 \]

\[ \text{Horizontal asymptote: } y=1 \]

3. First Derivative and Extrema

Given:

\[ h_1'(x)=\frac{-2e^{-x}(x-3)-2e^{-x}}{(x-3)^2} \]

Factor the numerator:

\[ -2e^{-x}(x-3)-2e^{-x} = -2e^{-x}(x-2) \]

Since
- \(e^{-x} > 0\)
- \((x-3)^2 > 0\)

the sign of \(h_1'(x)\) is determined by \(-(x-2)\).

Sign analysis - For \(x < 2\): \(x-2<0\) ⇒ numerator \(>0\) ⇒ increasing - For \(x > 2\): \(x-2>0\) ⇒ numerator \(<0\) ⇒ decreasing

6.12.1 Conclusion

\[ \text{Local maximum at } x = 2 \]

4. Second Derivative and Concavity

Start from:

\[ h_1'(x)=\frac{-2e^{-x}(x-2)}{(x-3)^2} \]

Let
\(N(x) = -2e^{-x}(x-2)\),
\(D(x) = (x-3)^2\).

Compute:

\[ N'(x) = 2e^{-x}(x-3) \] \[ D'(x) = 2(x-3) \]

Apply the quotient rule:

\[ h_1''(x)=\frac{2e^{-x}(x-3)(x-3)^2 + 2e^{-x}(x-2)\cdot 2(x-3)}{(x-3)^4} \]

Factor and simplify:

\[ h_1''(x)= \frac{2e^{-x}\left[(x-3)^2 + 2(x-2)\right]}{(x-3)^3} \]

Simplify numerator:

\[ (x-3)^2 + 2(x-2) = x^2 -4x +5 = (x-2)^2 + 1 > 0 \]

Since the numerator is always positive, the sign of \(h_1''(x)\) is controlled by the denominator \((x-3)^3\).

Concavity - For \(x<3\): \((x-3)^3 < 0\) ⇒ \(h_1''(x)<0\) ⇒ concave down - For \(x>3\): \((x-3)^3 > 0\) ⇒ \(h_1''(x)>0\) ⇒ concave up

Final Summary

Vertical asymptote: \(x=3\)
Horizontal asymptote: \(y=1\)
Increasing: \((-\infty, 2)\)
Decreasing: \((2, \infty)\)
Local max: at \(x=2\)
Concave down: \((-\infty, 3)\)
Concave up: \((3, \infty)\)

✔ Function \(h_2(x)=\dfrac{5}{x^{2}-4}+2\)

Vertical asymptotes:
\[ x=-2,\quad x=2 \]

Horizontal asymptote:
\[ \lim_{x\to\pm\infty}2 = 2 \]

First derivative:

\[ h_2'(x)=\frac{-10x}{(x^{2}-4)^2} \]

Critical point at \(x=0\).

Sign chart:

\(x<0\): \(h_2'(x)>0\) → increasing
\(x>0\): decreasing

→ Local max at \(x=0\)

Second derivative:

\[ h_2''(x)=\frac{30x^2+40}{(x^2-4)^3} \]

Concavity:

\(x<-2\): concave up
\(-2<x<2\): concave down
\(x>2\): concave up

✔ Function \(h_3(x)=\dfrac{e^{x}}{x+1}-4\)

1. Vertical Asymptote

\[ x = -1 \]

2. Horizontal Asymptotes

As \(x\to -\infty\): \[ \frac{e^x}{x+1} \to 0 \quad\Rightarrow\quad h_3(x)\to -4 \]

Horizontal asymptote:
\[ y=-4 \]

As \(x\to +\infty\): \[ \frac{e^x}{x+1} \to +\infty \]

No right-side horizontal asymptote.

3. First Derivative & Sign Chart

\[ h_3'(x)=\frac{e^x x}{(x+1)^2} \]

Critical point where numerator = 0 → \(x=0\)
Undefined at \(x=-1\)

Sign of \(h_3'(x)\):

\(x<0\): \(x<0\Rightarrow h_3'<0\) → decreasing
\(x>0\): \(x>0\Rightarrow h_3'>0\) → increasing

\[ \boxed{\text{Local minimum at } x=0} \]

4. Second Derivative & Concavity

Start with
\[ h_3'(x)=e^x x(x+1)^{-2} \]

Differentiate:

\[ \boxed{ h_3''(x)=\frac{e^x(x^2+1)}{(x+1)^3} } \]

Concavity analysis

Numerator \(e^x(x^2+1) > 0\) always
Denominator sign depends on \((x+1)^3\)

For \(x < -1\): \((x+1)^3<0\) → \(h_3''<0\)
\[ \boxed{\text{Concave down on }(-\infty,-1)} \]

For \(x > -1\): \((x+1)^3>0\) → \(h_3''>0\)
\[ \boxed{\text{Concave up on }(-1,\infty)} \]

5. Summary of Behavior

Vertical asymptote: \(x=-1\)
Horizontal asymptote (left only): \(y=-4\)
Increasing/decreasing:
- Decreasing on \((-\infty,0)\)
- Increasing on \((0,\infty)\)
Local extremum:
- Local minimum at \(x=0\)
Concavity:
- Concave down on \((-\infty,-1)\)
- Concave up on \((-1,\infty)\)

6. Qualitative Sketch Description

Approaches \(y=-4\) as \(x\to -\infty\)
Falls toward \(-\infty\) near the vertical asymptote at \(x=-1\)
Rises from \(-\infty\) after the asymptote
Has a smooth, concave-up minimum at \(x=0\)
Then increases without bound for large \(x\)