Unit 5: Analytical Applications of Differentiation

Extrema on Closed Intervals and Critical Points

A major goal of Unit 5 is learning how derivatives let you guarantee and locate maximum and minimum values of a function. In calculus language, maximum/minimum values are called extrema (plural of extremum). You’ll work with two main kinds:

• Absolute (global) maximum/minimum: the highest/lowest value of $f(x)$ on a specified domain (often an interval). No other function value on the domain is higher/lower.
• Local (relative) maximum/minimum: a value of $f(x)$ that is highest/lowest compared to nearby points (not necessarily the whole interval).

The derivative helps because derivatives measure instantaneous change. At many “peaks” and “valleys,” the function stops increasing and starts decreasing (or vice versa), which often corresponds to a derivative of zero.

Local extrema and critical numbers

A critical number of $f$ is a number $c$ in the domain of $f$ such that either:

• $f'(c) = 0$, or
• $f'(c)$ does not exist.

Critical numbers matter because local maxima/minima can only occur at critical numbers (or at endpoints of an interval, if endpoints are included). If a function is smooth and has a local high or low point in the interior of its domain, the tangent line there is horizontal.

Be careful: “critical number” refers to an $x$-value, not a $y$-value.

Why $f'(c)$ might not exist

Even if a function is continuous, its derivative can fail to exist. Typical reasons:

• Corner or cusp (sharp turn)
• Vertical tangent line (slope becomes infinite)
• Discontinuity (though if it’s discontinuous at $c$, then $c$ is not even in the domain)

A classic example is $f(x) = |x|$. The graph has a corner at $x = 0$, so $f'(0)$ does not exist, but $x = 0$ is still a critical number.

The Extreme Value Theorem (EVT)

The Extreme Value Theorem is what turns “maybe there’s a max/min” into “there must be a max/min.”

Extreme Value Theorem (EVT): If $f$ is **continuous** on a **closed interval** $[a,b]$, then $f$ attains both an **absolute maximum** value and an **absolute minimum** value on $[a,b]$.

This theorem is powerful because it gives you a guarantee, but only under its conditions:

• The interval must be closed (includes endpoints).
• The interval must be bounded (finite length).
• The function must be continuous on the entire interval.

If any condition fails, absolute extrema might not exist.

What can go wrong if EVT conditions fail

• On an open interval $(a,b)$, the function might approach a max/min but never reach it.
• If the function is discontinuous, it might “jump over” the highest/lowest value.

Example idea (open interval):

$f(x) = x$

on

$(0,1)$

has no absolute maximum because values get arbitrarily close to $1$ but never equal $1$.

The closed interval method (candidate’s test)

When EVT applies, you know absolute extrema exist. To find them on $[a,b]$, you use the closed interval method (often called the candidate’s test):

1. Find all critical numbers of $f$ in $(a,b)$.
2. Include the endpoints $a$ and $b$.
3. Make a quick table of these candidate $x$-values.
4. Plug each candidate into your original function $f$ (not into $f'$).
5. Compare all resulting function values. The largest is the absolute maximum; the smallest is the absolute minimum.

Why endpoints matter: an absolute max/min on $[a,b]$ can occur at endpoints even when the derivative is not zero there. Endpoints are not “surrounded” by points on both sides, so the usual horizontal-tangent logic for local extrema does not apply.

Worked example: absolute extrema on a closed interval

Find the absolute max and min of $f(x) = x^3 - 3x^2 + 1$ on $[0,3]$.

Step 1: derivative and critical numbers.

$f'(x) = 3x^2 - 6x$

Set

$f'(x)=0$

$3x^2 - 6x = 0$

Factor:

$3x(x-2) = 0$

So critical numbers are $x = 0$ and $x = 2$. But for the closed interval method, we include only critical numbers in the interior $(0,3)$, so $x=2$ is the interior critical number. (We will still evaluate the endpoint $x=0$ separately as an endpoint.)

Step 2: evaluate at critical numbers and endpoints.

$f(0) = 1$

$f(2) = -3$

$f(3) = 1$

Step 3: compare values.

• Smallest value is $-3$ at $x=2$, so the absolute minimum is $-3$.
• Largest value is $1$, which occurs at both $x=0$ and $x=3$, so the absolute maximum is $1$ (attained at two points).

Notice how the absolute maximum occurred at endpoints here. If you only checked $f'(x)=0$, you could easily miss it.

Worked example: EVT guarantee vs not guaranteed

Consider

$f(x) = \frac{1}{x}$

on $[1,3]$. It is continuous on $[1,3]$, so EVT applies and absolute extrema must exist.

Now consider

$f(x) = \frac{1}{x}$

on $[-1,1]$. This function is not continuous at $x=0$, so EVT does not apply, and indeed it does not have an absolute maximum or minimum on that interval (it becomes unbounded near $0$).

Exam Focus

• Typical question patterns
  • “Find the absolute maximum and minimum values of $f$ on $[a,b]$” (requires checking endpoints and critical points).
  • “Does $f$ have an absolute maximum on $[a,b]$? Justify.” (often expects EVT conditions: continuity and closed interval).
  • “Find critical points of $f$” including where $f'$ is undefined.
• Common mistakes
  • Forgetting to evaluate endpoints when asked for absolute extrema on $[a,b]$.
  • Listing points where $f'(x)=0$ but the point is not in the domain (critical numbers must be in the domain).
  • Using EVT on an interval that is open or where the function is discontinuous.

The Mean Value Theorem and What Derivatives Guarantee

Derivatives connect local behavior (instantaneous slope) to global behavior (overall change). The Mean Value Theorem (MVT) is the key statement that makes that connection precise: it links average rate of change to instantaneous rate of change.

Average rate of change vs instantaneous rate of change

On an interval $[a,b]$, the **average rate of change** of $f$ is the slope of the secant line connecting the endpoints:

$\frac{f(b)-f(a)}{b-a}$

The instantaneous rate of change at $x=c$ is the derivative:

$f'(c)$

MVT says that, under the right conditions, there is at least one point where the instantaneous rate equals the average rate. In geometric terms, there must be some point in the interval where the slope of the tangent line equals the slope of the secant line.

Mean Value Theorem (MVT)

Mean Value Theorem: If $f$ is **continuous** on $[a,b]$ and **differentiable** on $(a,b)$, then there exists at least one number $c$ in $(a,b)$ such that:

$f'(c) = \frac{f(b)-f(a)}{b-a}$

Interpretation: somewhere between $a$ and $b$, the tangent line is parallel to the secant line connecting $(a,f(a))$ and $(b,f(b))$.

The conditions matter:

• Continuous on $[a,b]$ ensures the graph has no jumps or holes.
• Differentiable on $(a,b)$ ensures no corners/cusps/vertical tangents inside.

If either condition fails, the conclusion may be false.

Rolle’s Theorem (special case of MVT)

Rolle’s Theorem: If $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, then there exists $c$ in $(a,b)$ such that:

$f'(c)=0$

This is MVT with average slope zero:

$\frac{f(b)-f(a)}{b-a} = 0$

Geometrically, if you start and end at the same height and you travel smoothly, you must have at least one point where you are “flat.” In other words, a continuous, differentiable curve with equal endpoint values has a horizontal tangent somewhere between those endpoints.

Why MVT matters in function analysis

MVT is not just a “find $c$” theorem. It supports big conclusions like:

• If $f'(x)=0$ for all $x$ in an interval, then $f$ is constant on that interval.
• If $f'(x)>0$ for all $x$ in an interval, then $f$ is increasing on that interval.
• If $f'(x)<0$ for all $x$ in an interval, then $f$ is decreasing on that interval.

These conclusions are used constantly in calculus proofs and in AP-style justifications.

Worked example: using MVT to find a point $c$

Let $f(x)=x^2$ on $[1,4]$. Find a value $c$ that satisfies MVT.

First check conditions: $x^2$ is continuous on $[1,4]$ and differentiable on $(1,4)$.

Compute average slope:

$\frac{f(4)-f(1)}{4-1} = 5$

Compute derivative:

$f'(x)=2x$

Set derivative equal to average slope:

$2c = 5$

So:

$c = \frac{5}{2}$

That value lies in $(1,4)$, so it works.

Worked example: when MVT does not apply

Let $f(x)=|x|$ on $[-1,1]$.

• Continuous on $[-1,1]$: yes.
• Differentiable on $(-1,1)$: no, because it’s not differentiable at $x=0$.

So MVT does not apply. The average slope is:

$\frac{f(1)-f(-1)}{1-(-1)} = 0$

MVT would claim there exists $c$ with $f'(c)=0$. But $f'(x)=-1$ for $x<0$ and $f'(x)=1$ for $x>0$, and at $0$ it does not exist. So there is no such $c$.

This is exactly why the differentiability condition is necessary.

Exam Focus

• Typical question patterns
  • “Verify that MVT applies on $[a,b]$ and find all $c$ that satisfy the conclusion.”
  • “Explain why MVT cannot be applied” (expects you to name the failed condition: continuity on $[a,b]$ or differentiability on $(a,b)$).
  • “Use MVT to justify that $f$ is increasing/decreasing/constant.”
• Common mistakes
  • Checking differentiability on $[a,b]$ instead of only on $(a,b)$ (endpoints don’t matter for differentiability in MVT).
  • Forgetting to verify conditions before solving for $c$.
  • Solving $f'(c)=\frac{f(b)-f(a)}{b-a}$ but giving a $c$ outside $(a,b)$.

Using the First Derivative to Describe Increasing, Decreasing, and Local Extrema

Once you can compute derivatives, the next step is learning how to read behavior from them. The first derivative tells you how the function is changing: positive derivative means the function is increasing; negative means decreasing.

Increasing and decreasing behavior

A function $f$ is **increasing** on an interval if, as $x$ increases, $f(x)$ increases. It is **decreasing** on an interval if, as $x$ increases, $f(x)$ decreases.

The derivative gives a precise test:

• If $f'(x) > 0$ on an interval, then $f$ is increasing on that interval.
• If $f'(x) < 0$ on an interval, then $f$ is decreasing on that interval.

Why this makes sense: $f'(x)$ is the slope of the tangent line. If slopes are positive everywhere, your graph is consistently rising.

A subtle but important point: knowing $f'(c)=0$ at one point does not tell you increasing/decreasing by itself. You need the sign of $f'$ in an interval around the point.

A reliable procedure for intervals of increase/decrease

When you’re given a formula for $f(x)$ and asked where it increases/decreases, a standard workflow is:

1. Take the derivative $f'(x)$.
2. Set $f'(x)=0$ (and also note where $f'(x)$ is undefined) to find critical numbers.
3. Break the number line into intervals using those critical numbers.
4. Pick a test value in each interval and plug it into $f'(x)$ to determine whether $f'(x)$ is positive or negative.

For instance, if $x=1$ is a critical number, you might test a number less than 1 (like 0) and a number greater than 1 (like 2). If $f'(0) > 0$, then $f$ is increasing to the left of 1; if $f'(2) < 0$, then $f$ is decreasing to the right of 1.

Local maxima and minima via sign changes (First Derivative Test idea)

A local maximum at $x=c$ means $f(c)$ is greater than nearby values. A typical pattern is that $f'$ changes from positive to negative at $c$.

A local minimum at $x=c$ typically corresponds to $f'$ changing from negative to positive at $c$.

These sign-change ideas are the heart of the First Derivative Test.

The First Derivative Test

Suppose $c$ is a critical number of $f$.

• If $f'(x) > 0$ for $x<c$ (near $c$) and $f'(x) < 0$ for $x>c$ (near $c$), then $f$ has a **local maximum** at $c$.
• If $f'(x) < 0$ for $x<c$ and $f'(x) > 0$ for $x>c$, then $f$ has a **local minimum** at $c$.
• If $f'$ does not change sign, then $f$ has **no local extremum** at $c$ (it could be a “flat” point where the function keeps increasing or keeps decreasing).

A good mental image is walking along a hill: if your slope changes from uphill to downhill, you passed a peak.

Worked example: intervals of increase/decrease and local extrema

Let $f(x) = x^3 - 3x$. Find where $f$ is increasing/decreasing and identify local extrema.

Step 1: compute derivative.

$f'(x) = 3x^2 - 3$

Step 2: critical numbers.

$3x^2 - 3 = 0$

$x^2 - 1 = 0$

$x = -1, 1$

Step 3: analyze sign of $f'(x)$.

$f'(x)=3(x-1)(x+1)$

Test intervals:

• For $x<-1$ (say $x=-2$), $f'(-2)>0$ so increasing on $(-\infty,-1)$.
• For $-1<x<1$ (say $x=0$), $f'(0)<0$ so decreasing on $(-1,1)$.
• For $x>1$ (say $x=2$), $f'(2)>0$ so increasing on $(1,\infty)$.

Step 4: local extrema via sign changes.

• At $x=-1$, $f'$ changes from positive to negative, so local maximum at $x=-1$.
• At $x=1$, $f'$ changes from negative to positive, so local minimum at $x=1$.

Local max/min values:

$f(-1) = 2$

$f(1) = -2$

Worked example: increasing/decreasing for a cubic with a constant term

Let

$f(x) = x^3 - 6x^2 + 9x + 2$

Find where the function is increasing or decreasing.

Different constants shift a graph up or down but do not change $f'(x)$, so the increasing/decreasing intervals come entirely from the derivative.

Compute the derivative:

$f'(x) = 3x^2 - 12x + 9$

Set to zero:

$3x^2 - 12x + 9 = 0$

Divide by 3:

$x^2 - 4x + 3 = 0$

Factor:

$(x-1)(x-3)=0$

So the critical numbers are $x=1$ and $x=3$.

Now test the sign of $f'(x)=3(x-1)(x-3)$:

• If $x<1$ (say $x=0$), then $f'(0)>0$, so $f$ is **increasing** on $(-\infty,1)$.
• If $1<x<3$ (say $x=2$), then $f'(2)<0$, so $f$ is **decreasing** on $(1,3)$.
• If $x>3$ (say $x=4$), then $f'(4)>0$, so $f$ is **increasing** on $(3,\infty)$.

Therefore, $f(x)$ is increasing on $(-\infty,1)$ and $(3,\infty)$, and decreasing on $(1,3)$.

From the sign changes, this function has a relative (local) maximum at $x=1$ (positive to negative) and a **relative (local) minimum** at $x=3$ (negative to positive).

A common “false positive”: critical point that is not an extremum

Consider $f(x)=x^3$.

$f'(x)=3x^2$

Critical number at $x=0$ because $f'(0)=0$. But $f'(x)=3x^2$ is nonnegative everywhere and positive for $x\ne 0$, so $f$ is increasing everywhere. The point $x=0$ is not a local max/min; it is a point where the tangent is horizontal but the function keeps increasing. (Later in this unit, you’ll connect this to concavity and inflection points.)

Exam Focus

• Typical question patterns
  • “Find intervals where $f$ is increasing/decreasing” (requires sign of $f'$).
  • “Find and classify local extrema” using the first derivative test.
  • “Given a graph of $f'$, describe $f$” (increasing where $f'$ is above the axis).
• Common mistakes
  • Assuming $f'(c)=0$ automatically means a max or min (you need a sign change).
  • Mixing up the meaning of $f'(x)>0$ (it means $f$ increases, not that $f(x)>0$).
  • Checking only one test point and forgetting that sign might differ on different intervals.

Concavity, the Second Derivative, and Inflection Points

The first derivative tells you whether the function is increasing or decreasing. The second derivative tells you how the rate of change itself is changing. This is where ideas like “bending upward” or “bending downward” become precise, and it explains phrases like “increasing, but at a decreasing rate.”

What concavity means (graph shape, not just slope)

A function is concave up on an interval if its graph bends like a cup: as you move left to right, the slopes tend to increase (they become less negative, then positive, then more positive). A function can be increasing and concave up (increasing faster), but it can also be increasing and concave down (still increasing, just at a decreasing rate).

A function is concave down on an interval if its graph bends like a frown: as you move left to right, the slopes tend to decrease.

Concavity is about the behavior of $f'$:

• Concave up means $f'$ is increasing.
• Concave down means $f'$ is decreasing.

The second derivative and concavity test

The second derivative $f''(x)$ is the derivative of the derivative:

$f''(x) = \frac{d}{dx}(f'(x))$

A standard concavity test is:

• If $f''(x) > 0$ on an interval, then $f$ is concave up on that interval.
• If $f''(x) < 0$ on an interval, then $f$ is concave down on that interval.

Why this makes sense: $f''(x)$ measures how the slope $f'(x)$ changes. If $f''(x)>0$, slopes are increasing, so the curve bends upward.

Points of inflection

A point of inflection is a point on the graph where the concavity changes (from up to down or down to up).

A common way to find candidates is to solve $f''(x)=0$ and also check where $f''(x)$ does not exist. However, a point is an inflection point only if concavity actually changes sign there.

A practical sign-chart process is:

1. Take the second derivative $f''(x)$.
2. Set $f''(x)=0$ (and note where $f''$ is undefined) to find candidate $x$-values.
3. Test a point on each side of each candidate to see whether $f''$ changes sign.

The Second Derivative Test (for local extrema)

The Second Derivative Test is a shortcut for classifying a critical point when $f'(c)=0$.

If $f'(c)=0$ and $f''(c)$ exists:

• If $f''(c) > 0$, $f$ has a local minimum at $c$.
• If $f''(c) < 0$, $f$ has a local maximum at $c$.
• If $f''(c) = 0$, the test is inconclusive.

A key limitation is that $f''(c)=0$ does not tell you anything by itself. You have to do more analysis.

Worked example: concavity and inflection points

Let $f(x)=x^4-2x^2$.

Step 1: compute derivatives.

$f'(x)=4x^3-4x$

$f''(x)=12x^2-4$

Step 2: concavity intervals.

Solve:

$12x^2-4=0$

$x=\pm \frac{1}{\sqrt{3}}$

Test intervals for the sign of $f''(x)=12x^2-4$:

• For $x=0$, $f''(0)<0$ so concave down on

$\left(-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)$

• For $x=1$, $f''(1)>0$ so concave up on

$\left(\frac{1}{\sqrt{3}},\infty\right)$

By symmetry it is also concave up on

$\left(-\infty,-\frac{1}{\sqrt{3}}\right)$

Step 3: inflection points.

Concavity changes at $x=\pm \frac{1}{\sqrt{3}}$, so those are inflection points. Find the corresponding $y$-values:

$f\left(\frac{1}{\sqrt{3}}\right)=\frac{1}{9}-\frac{2}{3}=-\frac{5}{9}$

So the inflection points are:

$\left(-\frac{1}{\sqrt{3}},-\frac{5}{9}\right)$

and

$\left(\frac{1}{\sqrt{3}},-\frac{5}{9}\right)$

Worked example: second derivative test (and when it fails)

Let $g(x)=x^3-3x$. Critical numbers are $x=-1$ and $x=1$.

$g''(x)=6x$

$g''(-1)<0$ so local maximum at $x=-1$.

$g''(1)>0$ so local minimum at $x=1$.

Now compare with $h(x)=x^3$:

$h'(x)=3x^2$

$h''(x)=6x$

At $x=0$, $h'(0)=0$ and $h''(0)=0$, so the second derivative test is inconclusive. In fact, $x=0$ is an inflection point, not a max or min.

Worked example: concavity for a polynomial used earlier

Suppose

$f(x) = x^3 - 6x^2 + 9x + 2$

Find where the function is concave up or concave down.

Compute derivatives:

$f'(x) = 3x^2 - 12x + 9$

$f''(x) = 6x - 12$

Set the second derivative equal to zero:

$6x - 12 = 0$

$x = 2$

Test the sign of $f''(x)$:

• If $x<2$ (say $x=1$), then $f''(1)<0$ so $f$ is concave down on $(-\infty,2)$.
• If $x>2$ (say $x=3$), then $f''(3)>0$ so $f$ is concave up on $(2,\infty)$.

Because the sign changes at $x=2$, $x=2$ is an inflection point (and on an actual free-response question you’d typically report the coordinate by also finding $f(2)$).

Exam Focus

• Typical question patterns
  • “Find intervals of concavity and points of inflection” (requires sign analysis of $f''$).
  • “Use the second derivative test to classify critical points.”
  • “Given a graph of $f''$, determine where $f$ is concave up/down.”
• Common mistakes
  • Declaring an inflection point whenever $f''(c)=0$ (concavity must actually change).
  • Confusing concavity with increasing/decreasing (concave up does not necessarily mean increasing).
  • Using the second derivative test when $f'(c)\ne 0$ (it only applies at critical points where $f'(c)=0$).

Connecting $f$, $f'$, and $f''$ Through Graphs and Tables

AP Calculus frequently asks you to move between a function and its derivatives without doing heavy algebra. You might be given a graph of $f$ and asked about $f'$, or given a graph of $f'$ and asked to sketch $f$. The key is to translate each graph feature into a derivative meaning.

A quick notation reference

You’ll see derivatives written in several equivalent ways:

Reading $f'$ from the graph of $f$

From a graph of $f$:

• Where $f$ is increasing, $f'(x) > 0$.
• Where $f$ is decreasing, $f'(x) < 0$.
• Where $f$ has a horizontal tangent (often at a local extremum), $f'(x) = 0$.
• Where $f$ has a corner/cusp/vertical tangent, $f'(x)$ may not exist.

Steepness matters: a very steep positive slope means $f'(x)$ is a large positive number.

Reading concavity and $f''$ from the graph of $f$

From a graph of $f$:

• Concave up means $f''(x) > 0$.
• Concave down means $f''(x) < 0$.
• Inflection point means $f''$ changes sign.

A useful visual cue is to track tangent slopes: if the tangents are getting steeper (slopes increasing), concave up; if the tangents are getting less steep (slopes decreasing), concave down.

Reading $f$ from a graph of $f'$

From a graph of $f'$:

• Where $f'$ is above the $x$-axis (positive), $f$ is increasing.
• Where $f'$ is below the $x$-axis (negative), $f$ is decreasing.
• Where $f'$ crosses the $x$-axis (changes sign), $f$ has a local extremum.
  • Positive to negative crossing: local maximum.
  • Negative to positive crossing: local minimum.
• Where $f'$ is increasing, $f$ is concave up (because $f''>0$).
• Where $f'$ is decreasing, $f$ is concave down (because $f''<0$).

Reading $f'$ from a table of values of $f$

Sometimes you are given a table of function values and asked to estimate or determine signs.

• You can approximate average rate of change on subintervals:

$\frac{f(x_2)-f(x_1)}{x_2-x_1}$

Those slopes approximate $f'$ at points between $x_1$ and $x_2$ (MVT logic).

• If the table shows $f$ values rising as $x$ increases, that suggests $f'$ is positive over that region.

Be cautious: tables give limited data, so questions often ask for conclusions supported by the information (like sign or an estimate), not an exact derivative formula.

Worked example: describing $f$ given a sign chart for $f'$

Suppose you know:

• $f'(x) < 0$ on $(-2,0)$
• $f'(x) > 0$ on $(0,3)$
• $f'(x) < 0$ on $(3,5)$

Then:

• $f$ is decreasing on $(-2,0)$.
• $f$ is increasing on $(0,3)$.
• $f$ is decreasing on $(3,5)$.

Local extrema:

• At $x=0$, $f'$ changes from negative to positive, so $f$ has a local minimum at $x=0$.
• At $x=3$, $f'$ changes from positive to negative, so $f$ has a local maximum at $x=3$.

Worked example: relating concavity of $f$ to monotonicity of $f'$

If you are told $f'(x)$ is increasing on $(-1,4)$, then $f''(x) > 0$ on $(-1,4)$ (at least where $f''$ exists), and $f$ is concave up on $(-1,4)$.

A common confusion to avoid: “$f'$ increasing” does **not** mean “$f$ increasing.” “$f'$ increasing” means slopes are increasing, which is concavity information.

Exam Focus

• Typical question patterns
  • “Given the graph of $f$, sketch or describe the graph of $f'$.”
  • “Given the graph of $f'$, identify intervals where $f$ is increasing/decreasing and where $f$ has local extrema.”
  • “Use a table/graph of $f'$ to determine concavity of $f$ and locate inflection points.”
• Common mistakes
  • Confusing the value of $f$ with the value of $f'$ (a high $f(x)$ does not mean a high slope).
  • Saying $f$ has a local extremum wherever $f'(x)=0$ even if $f'$ does not change sign.
  • Mixing up “concave up” with “increasing” (a function can be decreasing and concave up at the same time).

Sketching a Function Using Derivative Information

In many calculus problems you are not asked for an exact graph; instead, you’re asked to create a qualitative sketch based on derivative information. This is a powerful analytical skill: you turn algebraic or derivative facts into a coherent picture.

What a derivative-based sketch should show

A good calculus sketch (even if it’s rough) should reflect:

• Intercepts or key points that you can compute.
• Where the function increases/decreases (from $f'$).
• Where it is concave up/down (from $f''$).
• Locations of local maxima/minima (from critical points and sign change of $f'$).
• Locations of inflection points (from sign change of $f''$).

You’re not trying to “connect the dots randomly.” You’re building a graph that is consistent with the derivative constraints.

A process you can rely on

When you’re given $f(x)$ and asked to sketch using calculus, a typical workflow is:

1. Compute $f'(x)$ and $f''(x)$.
2. Find critical numbers from $f'(x)=0$ or where $f'$ is undefined.
3. Use the sign of $f'(x)$ to determine increasing/decreasing and local extrema.
4. Find possible inflection points from $f''(x)=0$ or where $f''$ is undefined.
5. Use the sign of $f''(x)$ to determine concavity and confirm inflection points.
6. Evaluate $f(x)$ at key $x$-values (critical points, inflection points, intercepts) to plot anchor points.
7. Draw a smooth curve consistent with all the above.

On the AP Exam, you might not be asked to compute all of these from scratch; sometimes you’re given a graph of $f'$ or a table. But the logic is the same.

Worked example: full analysis and sketch features

Analyze $f(x)=x^3-6x^2+9x+1$: find local extrema, concavity, and inflection point.

Step 1: first derivative and critical points.

$f'(x)=3x^2-12x+9$

Set $f'(x)=0$:

$3x^2-12x+9=0$

Divide by 3:

$x^2-4x+3=0$

Factor:

$(x-1)(x-3)=0$

So critical numbers are $x=1$ and $x=3$.

Step 2: increasing/decreasing via sign of $f'$.

Since

$f'(x)=3(x-1)(x-3)$

Test intervals:

• For $x<1$, choose $x=0$: $f'(0)>0$ so increasing on $(-\infty,1)$.
• For $1<x<3$, choose $x=2$: $f'(2)<0$ so decreasing on $(1,3)$.
• For $x>3$, choose $x=4$: $f'(4)>0$ so increasing on $(3,\infty)$.

Local extrema:

• At $x=1$, $f'$ changes positive to negative, so local maximum at $x=1$.
• At $x=3$, $f'$ changes negative to positive, so local minimum at $x=3$.

Compute the values:

$f(1)=5$

$f(3)=1$

So local max at $(1,5)$ and local min at $(3,1)$.

Step 3: concavity via second derivative.

$f''(x)=6x-12$

Set $f''(x)=0$:

$6x-12=0$

$x=2$

Test concavity:

• For $x<2$, say $x=1$: $f''(1)<0$ so concave down on $(-\infty,2)$.
• For $x>2$, say $x=3$: $f''(3)>0$ so concave up on $(2,\infty)$.

So there is an inflection point at $x=2$. Compute:

$f(2)=3$

Inflection point at $(2,3)$.

What your sketch should reflect:

• Increasing up to $x=1$, then decreasing until $x=3$, then increasing afterward.
• Concave down until $x=2$, then concave up afterward.
• A local max at $(1,5)$.
• An inflection point at $(2,3)$.
• A local min at $(3,1)$.

Even without perfect scaling, a sketch consistent with these facts will look “cubic-like,” rising left to right, with a peak, then a dip, and concavity changing once.

Common sketching pitfalls

• Drawing a local maximum at a point where $f'(x)=0$ but $f'$ does not change sign.
• Forgetting that concavity and increasing/decreasing are independent (you can be decreasing and concave up).
• Treating inflection points like “turning points.” An inflection point is about concavity change, not about switching from increasing to decreasing.

Exam Focus

• Typical question patterns
  • “Use derivatives to sketch the graph of $f$” (expects critical points, monotonicity, concavity, inflection points).
  • “Given a graph of $f'$, sketch a possible $f$” (many answers possible; must be consistent).
  • “Find the x-coordinates of inflection points/local extrema and justify using sign changes.”
• Common mistakes
  • Reporting critical numbers as points on the graph without computing $f(c)$ (AP often wants coordinates).
  • Claiming “inflection point” at every solution to $f''(x)=0$ without checking sign change.
  • Overfocusing on algebra and not using sign reasoning (many questions are about interpretation).