Unit 5 Curve Sketching and Optimization: Using Derivatives to Understand and Optimize Functions

Connecting $f$, $f'$, and $f''$ Graphs

What these three graphs represent (and why you should care)

When you study a function $f$ in calculus, you are often not just interested in its exact formula—you want to understand its _behavior_: where it rises or falls, where it levels off, where it bends upward or downward, and where it reaches high or low points. The graphs of $f$, its first derivative $f'$, and its second derivative $f''$ are three “views” of the same object.

• The graph of $f$ shows the output values (heights) of the function.
• The graph of $f'$ shows the _slope_ of $f$ at each $x$ value. It tells you how fast $f$ is changing and whether $f$ is increasing or decreasing.
• The graph of $f''$ shows how the slope is changing. It tells you about **concavity**—whether $f$ bends like a cup up or a cup down.

This matters because many AP Calculus AB questions give you one of these graphs (or partial information about one) and ask you to infer properties of the others. If you can translate “shape language” between these graphs, you can solve problems without heavy algebra.

The core connections

1) Increasing/decreasing and the sign of $f'$

The derivative $f'(x)$ is the slope of the tangent line to $f$ at $x$.

• If $f'(x) > 0$, tangent lines slope upward left-to-right, so $f$ is increasing.
• If $f'(x) < 0$, tangent lines slope downward, so $f$ is decreasing.
• If $f'(x) = 0$, the tangent line is horizontal; $x$ is a critical point (a candidate for a local max/min, but not guaranteed).

A key translation between graphs:

• Where $f$ has a horizontal tangent, the graph of $f'$ hits $0$.

2) Local extrema of $f$ and zeros/sign-changes of $f'$

A local maximum of $f$ occurs when $f$ changes from increasing to decreasing. A **local minimum** occurs when $f$ changes from decreasing to increasing.

On the derivative graph, that means:

• Local max of $f$ at $x=c$ often corresponds to $f'(c)=0$ with $f'$ changing from positive to negative.
• Local min of $f$ at $x=c$ often corresponds to $f'(c)=0$ with $f'$ changing from negative to positive.

Important: $f'(c)=0$ alone does **not** guarantee an extremum. If $f'$ does not change sign (for example, it touches 0 and stays positive), then $f$ may just flatten briefly.

3) Concavity of $f$ and the sign of $f''$

Concavity describes how the slope of $f$ is changing.

• If $f''(x) > 0$, slopes of $f$ are increasing (becoming more positive or less negative), so $f$ is concave up.
• If $f''(x) < 0$, slopes of $f$ are decreasing, so $f$ is concave down.

A very useful way to think about this: concavity is about whether tangent lines are “rotating” upward or downward as you move to the right.

4) Inflection points of $f$ and zeros/sign-changes of $f''$

An inflection point is where $f$ changes concavity (concave up to concave down, or vice versa).

• Typically, $f''(c)=0$ (or does not exist) at an inflection point, **and** $f''$ changes sign around $c$.

As with extrema, $f''(c)=0$ alone is not enough—you need the concavity change.

5) Peaks/valleys of $f'$ correspond to inflection points of $f$

Because $f''$ is the derivative of $f'$:

• Local maxima/minima of $f'$ occur where $f''(x)=0$ with a sign change.
• Those same $x$-values are often where $f$ has inflection points (because slope stops increasing and starts decreasing, or vice versa).

A relationship table you can translate quickly

Example 1: Inferring behavior of $f$ from a sign chart of $f'$ and $f''$

Suppose you are told:

• $f'(x) > 0$ on $(-2, 1)$ and $f'(x) < 0$ on $(1, 4)$.
• $f''(x) < 0$ on $(-2, 0)$ and $f''(x) > 0$ on $(0, 4)$.

What can you conclude about $f$?

1) Increasing/decreasing:

• $f$ is increasing on $(-2, 1)$.
• $f$ is decreasing on $(1, 4)$.
  So $f$ has a **local maximum** at $x=1$ (because $f'$ changes from positive to negative).

2) Concavity:

• $f$ is concave down on $(-2, 0)$.
• $f$ is concave up on $(0, 4)$.
  So $f$ has an **inflection point** at $x=0$ (concavity changes).

Notice what you cannot conclude: you cannot find exact $y$-values of the max or inflection point without more information about $f$.

Example 2: Matching a feature across graphs

If the graph of $f$ has a point where the tangent line is steepest (largest slope) at $x=c$, that means $f'(x)$ is largest at $x=c$—so $f'$ has a local maximum there. Then $f''(c)=0$ (typically) with a sign change from positive to negative.

This “steepest slope” language is common: it’s really a question about extrema of $f'$.

Exam Focus

• Typical question patterns:
  • Given a graph of $f$, identify intervals where $f' > 0$, $f' < 0$, and where $f'' > 0$, $f'' < 0$.
  • Given sign charts or partial graphs for $f'$ and/or $f''$, determine where $f$ has local extrema and inflection points.
  • Match unlabeled graphs of $f$, $f'$, and $f''$ by comparing zeros, sign changes, and “where slope is increasing/decreasing.”
• Common mistakes:
  • Treating $f'(c)=0$ as “definitely a max/min” without checking for a sign change in $f'$.
  • Calling any point where $f''(c)=0$ an inflection point without verifying concavity changes.
  • Confusing “concave up” with “increasing”—a function can be decreasing and concave up at the same time.

Sketching Graphs of Functions and Their Derivatives

The goal: turning derivative information into a shape

“Curve sketching” in AP Calculus AB usually does not mean producing a perfect artistic graph. It means you can build a mathematically justified sketch using key features: intercepts, asymptotes (if relevant), intervals of increase/decrease, local extrema, concavity, and inflection points.

The central idea is that derivatives convert shape into sign information:

• $f'$ controls up/down behavior.
• $f''$ controls bending behavior.

You’ll do this in two directions:
1) Use algebraic expressions for $f'$ and $f''$ to sketch $f$.
2) Use a graph of $f$ to sketch qualitative graphs of $f'$ and sometimes $f''$.

Sketching $f$ from a formula: a structured process

When you are given a function $f(x)$, the sketching workflow is basically a conversation with the function: “Where are you defined? Where do you cross axes? Where do you flatten? Where do you bend?”

Step 1: Domain and discontinuities

Before using derivatives, identify where $f$ is defined. Points where $f$ is undefined can create vertical asymptotes or holes, and they break the number line into intervals.

For rational functions, solve where the denominator is zero. For roots, ensure the radicand is nonnegative (for even roots). These domain breaks matter because sign charts for $f'$ and $f''$ must respect them.

Step 2: Intercepts and basic anchor points

Compute:

• $y$-intercept: evaluate $f(0)$ (if $0$ is in the domain).
• $x$-intercepts: solve $f(x)=0$ if feasible.

Even a rough sketch becomes much easier when you have a few fixed points.

Step 3: First derivative gives monotonicity and local extrema

Compute $f'(x)$, then find critical numbers where:

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

Then test the sign of $f'$ on each interval to determine where $f$ increases or decreases.

A common AP tool is the First Derivative Test:

• If $f'$ changes from positive to negative at $x=c$, $f$ has a local maximum at $x=c$.
• If $f'$ changes from negative to positive at $x=c$, $f$ has a local minimum at $x=c$.

Step 4: Second derivative gives concavity and inflection points

Compute $f''(x)$ and find where $f''(x)=0$ or does not exist (again, where $f$ is still defined). Then test the sign of $f''$ to determine concavity.

The Second Derivative Test is a shortcut that sometimes applies at critical points where $f'(c)=0$:

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

Why it works: $f''$ measures whether slope is increasing or decreasing near the critical point.

Example 1: Sketching using $f'$ and $f''$

Let:
$f(x) = x^3 - 3x$

1) Compute derivatives:
$f'(x) = 3x^2 - 3$
$f''(x) = 6x$

2) Critical numbers from $f'(x)=0$:
$3x^2 - 3 = 0$
$x^2 = 1$
$x = -1, 1$

3) Sign of $f'$:
Because $f'(x)=3(x^2-1)$:

• For $x < -1$, $x^2 - 1 > 0$ so $f' > 0$ (increasing).
• For $-1 < x < 1$, $x^2 - 1 < 0$ so $f' < 0$ (decreasing).
• For $x > 1$, $f' > 0$ (increasing).

So $f$ has a local maximum at $x=-1$ and a local minimum at $x=1$.

4) Concavity from $f''(x)=6x$:

• For $x < 0$, $f'' < 0$, so concave down.
• For $x > 0$, $f'' > 0$, so concave up.
  Thus $f$ has an inflection point at $x=0$.

5) Anchor points (optional but helpful):
$f(0)=0$
$f(-1)=(-1)^3-3(-1)=2$
$f(1)=1-3=-2$

Now you can sketch: increasing to a local max at $(-1,2)$, decreasing through the origin with concave down until $x=0$, then concave up afterward, hitting a local min at $(1,-2)$, then increasing.

Sketching $f'$ from the graph of $f$ (no formula needed)

On the AP exam, you may be given a graph of $f$ and asked to sketch $f'$. The key is to remember that $f'(x)$ is the slope of $f$.

Here’s how you translate visually:

1) Where $f$ is increasing, the slope is positive, so $f'$ should be above the $x$-axis.

2) Where $f$ is decreasing, slope is negative, so $f'$ should be below the $x$-axis.

3) Where $f$ has a horizontal tangent, slope is zero, so $f'$ crosses (or touches) the $x$-axis.

4) Steepness matters: if $f$ is very steep upward, $f'$ should be a large positive value; if $f$ is flattening, $f'$ should approach zero.

5) Corners, cusps, and vertical tangents: if $f$ has a sharp corner (like an absolute value point), $f'$ typically does not exist there. On the sketch of $f'$, you show a break/open circle at that $x$.

Example 2: From a piecewise-linear $f$ to a step-like $f'$

If $f$ is made of straight-line segments, then each segment has constant slope, so $f'$ is constant on each interval.

For instance, if on $(-2,0)$ the graph of $f$ is a line rising with slope $2$, then $f'(x)=2$ on $(-2,0)$. If on $(0,3)$ the graph is a line falling with slope $-1$, then $f'(x)=-1$ on $(0,3)$. At $x=0$, if there is a corner, then $f'$ does not exist.

This is a powerful mental model: “curvy function” gives “smooth derivative,” while “straight segments” give “flat derivative.”

Sketching $f''$ from the graph of $f$ (and from $f'$)

If you already have a sketch of $f'$, then sketching $f''$ becomes the same task again: $f''$ is the slope of $f'$.

• Where $f'$ is increasing, $f'' > 0$.
• Where $f'$ is decreasing, $f'' < 0$.
• Where $f'$ has a local max/min, $f'' = 0$ (often).

Equivalently, from $f$ directly:

• If $f$ is concave up, $f'' > 0$.
• If $f$ is concave down, $f'' < 0$.

What can go wrong when sketching derivatives

A frequent mistake is to copy the shape of $f$ and call it $f'$. Instead, you must _re-encode_ the information: heights of $f'$ are slopes of $f$.

Another common pitfall: placing zeros of $f'$ at intercepts of $f$. Intercepts are where $f=0$, but zeros of $f'$ are where the slope is zero. Those are unrelated unless the graph happens to have horizontal tangents at intercepts.

Exam Focus

• Typical question patterns:
  • Given a graph of $f$, sketch $f'$ by identifying where slopes are positive/negative/zero and where $f'$ is undefined.
  • Given a graph of $f'$ (or a sign chart), sketch a possible graph of $f$ that matches the derivative information.
  • Use $f'$ and $f''$ (from formulas or tables) to justify local extrema and inflection points for a sketch.
• Common mistakes:
  • Confusing “$f(x)=0$” with “$f'(x)=0$” and placing critical points at intercepts.
  • Forgetting that derivative graphs may be undefined where the original function has corners or vertical tangents.
  • Mixing up concavity and monotonicity (e.g., claiming concave up means increasing).

Optimization Problems

What optimization means in calculus

An optimization problem asks you to find the best possible value of some quantity: maximum area, minimum cost, shortest distance, greatest volume, and so on. Calculus makes these problems systematic because many “best” values occur at places where the rate of change switches direction—exactly what derivatives detect.

In AP Calculus AB, optimization problems are usually single-variable calculus problems after you translate the situation into an objective function. The big skill is not the derivative itself—it’s building the correct function to optimize and using the correct domain.

Why derivatives locate maxima and minima

If a function $Q(x)$ represents the quantity you want to maximize or minimize, then interior extrema typically happen where the slope is zero:
$Q'(x)=0$

But absolute (global) maxima/minima on a closed interval can also occur at endpoints. That’s why the Closed Interval Method is central:
1) Find critical numbers where $Q'(x)=0$ or $Q'$ does not exist (and $Q$ does).
2) Evaluate $Q$ at each critical number in the interval.
3) Evaluate $Q$ at the interval endpoints.
4) Compare values to decide absolute max/min.

A subtle but important point: in word problems, the “interval” is the set of physically possible values (nonnegative lengths, feasible dimensions, etc.). Setting the domain correctly prevents nonsense answers.

A reliable setup process (how optimization works)

Step 1: Draw a picture and name variables

Even a simple sketch reduces errors. Label the variable you will ultimately differentiate with respect to, such as $x$.

Step 2: Write the objective function

Decide what you are optimizing. Examples:

• Area: $A$
• Volume: $V$
• Cost: $C$
• Distance/time: $D$ or $T$

Write it as a function, like $A = A(x)$.

Step 3: Use constraints to reduce to one variable

Most problems start with multiple variables (length and width, radius and height, etc.). Use given constraints (like fixed perimeter, fixed surface area, or geometric relationships) to rewrite everything in terms of one variable.

Step 4: State the domain

Use the context: lengths must be positive, radicands must be nonnegative, denominators nonzero, etc.

Step 5: Differentiate, solve, and interpret

Compute $A'(x)$ (or whatever derivative), solve $A'(x)=0$ in the domain, and use either the Closed Interval Method or a derivative test to confirm max/min. Then translate back to the requested quantities (not just $x$).

Example 1: Maximize area with fixed perimeter (classic rectangle)

A rectangle has perimeter $P=100$. What dimensions maximize the area?

1) Variables and constraint
Let length be $L$ and width be $W$.
Perimeter constraint:
$2L + 2W = 100$
So:
$L + W = 50$
$W = 50 - L$

2) Objective function
Area:
$A = LW$
Substitute:
$A(L) = L(50 - L)$

3) Domain
You need $L > 0$ and $W > 0$, so:
$0 < L < 50$

4) Differentiate and find critical point
$A(L) = 50L - L^2$
$A'(L) = 50 - 2L$
Set to zero:
$50 - 2L = 0$
$L = 25$
Then:
$W = 50 - 25 = 25$

5) Conclude
The rectangle with maximum area is a square: $25$ by $25$.

A common interpretation insight: when a constraint is symmetric (like fixed perimeter) and the objective is also symmetric (area treats length and width equally), the optimum often occurs when the variables are equal.

Example 2: Minimize surface area of a cylinder with fixed volume

A closed cylinder (has top and bottom) must have volume $V_0$. Find the radius-to-height relationship that minimizes surface area.

1) Variables and formulas
Let radius be $r$ and height be $h$.
Volume constraint:
$V = \pi r^2 h = V_0$
Surface area (top + bottom + side):
$S = 2\pi r^2 + 2\pi r h$

2) Reduce to one variable using the constraint
Solve the volume equation for $h$:
$h = \frac{V_0}{\pi r^2}$
Substitute into $S$:
$S(r) = 2\pi r^2 + 2\pi r\left(\frac{V_0}{\pi r^2}\right)$
Simplify:
$S(r) = 2\pi r^2 + \frac{2V_0}{r}$

3) Domain
$r > 0$.

4) Differentiate and solve
$S'(r) = 4\pi r - \frac{2V_0}{r^2}$
Set to zero:
$4\pi r - \frac{2V_0}{r^2} = 0$
$4\pi r = \frac{2V_0}{r^2}$
Multiply by $r^2$:
$4\pi r^3 = 2V_0$
$2\pi r^3 = V_0$

Now use the volume constraint again:
$V_0 = \pi r^2 h$
Set equal:
$\pi r^2 h = 2\pi r^3$
Cancel $\pi r^2$:
$h = 2r$

5) Interpret
The cylinder with minimum surface area for a fixed volume satisfies $h = 2r$ (height equals diameter).

Notice how the algebra was manageable only after reducing to one variable.

Example 3: Closest point on a curve (distance minimization)

Find the point on the curve
$y = x^2$
closest to the point $(0,1)$.

1) Choose what to minimize
Distance from $(x, x^2)$ to $(0,1)$ is:
$D = \sqrt{(x-0)^2 + (x^2-1)^2}$
Minimizing $D$ is equivalent to minimizing $D^2$ (because square root is increasing for nonnegative inputs):
$D^2 = x^2 + (x^2 - 1)^2$
This is a common trick because it avoids messy square roots.

2) Objective function
$Q(x) = x^2 + (x^2 - 1)^2$
Expand:
$(x^2 - 1)^2 = x^4 - 2x^2 + 1$
So:
$Q(x) = x^2 + x^4 - 2x^2 + 1 = x^4 - x^2 + 1$

3) Differentiate
$Q'(x) = 4x^3 - 2x$
Set to zero:
$4x^3 - 2x = 0$
Factor:
$2x(2x^2 - 1) = 0$
So:
$x = 0$
$2x^2 - 1 = 0$
$x^2 = \frac{1}{2}$
$x = \pm \frac{1}{\sqrt{2}}$

4) Compare values (absolute min over all real numbers)
Evaluate $Q$:

• $Q(0)=1$
• If $x^2=\frac{1}{2}$ then $x^4=\frac{1}{4}$, so:
  $Q = \frac{1}{4} - \frac{1}{2} + 1 = \frac{3}{4}$
  Thus the minimum occurs at:
  $x = \pm \frac{1}{\sqrt{2}}$
  Corresponding $y$ values:
  $y = x^2 = \frac{1}{2}$
  So the closest points are:
  $\left(\frac{1}{\sqrt{2}}, \frac{1}{2}\right), \left(-\frac{1}{\sqrt{2}}, \frac{1}{2}\right)$

Common traps in optimization (and how to avoid them)

• Optimizing the wrong function: For distance problems, minimizing $D$ vs minimizing $D^2$ gives the same $x$-location, but if you mix them partway through, you can make algebra mistakes. Choose one and stick with it.
• Forgetting endpoints: If the domain is a closed interval (common in geometry with nonnegative dimensions), you must check endpoints for absolute extrema.
• Losing the meaning of the variable: Often you solve for $x$ but the question asks for dimensions, a point, or a maximum value. Always translate back.
• Domain errors: You might find a critical point like $x=-3$, but if $x$ represents a length, it’s invalid.

Exam Focus

• Typical question patterns:
  • “Maximize/minimize area/volume/cost” with a geometric constraint (fixed perimeter, fixed area, given amount of material).
  • “Find dimensions that minimize cost” where different parts have different per-unit costs (leading to a cost function like $C(x)$).
  • Distance minimization (closest point) using $D^2$ to simplify.
• Common mistakes:
  • Differentiating before reducing to one variable, then getting stuck with multiple variables.
  • Finding a critical point but not verifying it’s in the feasible domain (or not comparing with endpoints).
  • Reporting only the critical $x$ value when the prompt asks for the optimized quantity (maximum area, minimum cost, or full dimensions).