Analytical Applications of Differentiation: Curve Sketching & Optimization

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

When you “connect” the graphs of $f$, $f'$, and $f''$, you’re translating between three different views of the same function:

• $f$ tells you position/output (the actual height of the graph).
• $f'$ tells you **rate of change** (the slope of $f$).
• $f''$ tells you **how the rate of change is changing** (concavity of $f$).

This matters because many AP questions don’t give you a formula for $f$. Instead, you might get a graph of $f'$ (or $f''$) and be asked to deduce where $f$ increases, where it has a local maximum, where it is concave up, and so on. That’s curve sketching in its “information-based” form.

What $f'$ tells you about $f$ (increasing/decreasing and extrema)

The derivative $f'$ measures the slope of $f$. If you imagine walking along the graph of $f$ from left to right:

• If the graph is going upward, slopes are positive, so $f' > 0$.
• If the graph is going downward, slopes are negative, so $f' < 0$.
• If the graph is flat (horizontal tangent), then $f' = 0$.

So the sign of $f'$ controls monotonicity:

• $f$ is increasing on intervals where $f' > 0$.
• $f$ is decreasing on intervals where $f' < 0$.

A critical point of $f$ is a point in the domain where $f' = 0$ or $f'$ does not exist. Critical points are important because local extrema (local maxima/minima) can only occur at critical points.

But a very common mistake is thinking “$f' = 0$ means max or min.” Not necessarily. A horizontal tangent could be a max, a min, or neither (like an S-shaped flattening).

To decide what happens at a critical point, you look for a sign change in $f'$:

• If $f'$ changes from positive to negative, $f$ has a local maximum.
• If $f'$ changes from negative to positive, $f$ has a local minimum.
• If $f'$ does not change sign, $f$ has no local extremum there.

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

Suppose $f'$ is positive on $(-\infty, 1)$, equals $0$ at $1$, is negative on $(1, 4)$, equals $0$ at $4$, and is positive on $(4, \infty)$.

• On $(-\infty, 1)$, $f' > 0$ so $f$ is increasing.
• At $x = 1$, $f'$ changes from $+$ to $-$, so $f$ has a local maximum at $x = 1$.
• On $(1, 4)$, $f' < 0$ so $f$ is decreasing.
• At $x = 4$, $f'$ changes from $-$ to $+$, so $f$ has a local minimum at $x = 4$.

Notice that you can sketch a reasonable shape of $f$ (up then down then up) without ever knowing the exact formula.

What $f''$ tells you about $f$ (concavity and inflection points)

Concavity is about how slopes are changing.

• Concave up means the slopes of $f$ are increasing as $x$ increases (tangent lines get steeper). This corresponds to $f'' > 0$.
• Concave down means the slopes of $f$ are decreasing as $x$ increases (tangent lines get less steep). This corresponds to $f'' < 0$.

So:

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

An inflection point of $f$ is a point where the concavity changes (from up to down, or down to up). A necessary condition is that $f'' = 0$ or $f''$ is undefined, but—again—this is not sufficient. You must confirm that $f''$ changes sign.

A helpful way to remember concavity is:

• Concave up: the graph of $f$ looks like a cup “holding water.”
• Concave down: the graph of $f$ looks like a cap “spilling water.”

Connecting $f''$ and $f'$ (increasing/decreasing of the derivative)

Since $f''$ is the derivative of $f'$, it tells you whether $f'$ is increasing or decreasing:

• If $f'' > 0$ on an interval, then $f'$ is increasing there.
• If $f'' < 0$ on an interval, then $f'$ is decreasing there.

This is a powerful connection on AP problems where you’re shown the graph of $f'$. You can often determine concavity of $f$ by checking whether $f'$ is increasing or decreasing.

• $f$ is concave up where $f'$ is increasing.
• $f$ is concave down where $f'$ is decreasing.

This works because “concavity of $f$” is really “trend in the slope of $f$.”

Example 2: Reading concavity of $f$ from a graph of $f'$

Imagine you’re given a graph of $f'$ that rises from left to right on the interval $(0, 2)$ and falls from left to right on $(2, 5)$.

• On $(0, 2)$, $f'$ is increasing, so $f'' > 0$ there and $f$ is concave up.
• On $(2, 5)$, $f'$ is decreasing, so $f'' < 0$ there and $f$ is concave down.

If $f'$ changes from increasing to decreasing at $x = 2$, that suggests $f''$ changes sign, so $f$ likely has an inflection point at $x = 2$ (as long as $f$ is continuous there).

How to sketch $f$ when you’re given $f'$ (a practical workflow)

On many free-response questions, you’re given a graph of $f'$ and asked to sketch a graph of $f$ that is consistent with it. You’re not expected to find the exact formula—just a graph that matches the derivative information.

A reliable approach:

1. Mark where $f' = 0$ or undefined. These are candidate points where $f$ could have local extrema or corners/cusps.
2. Determine increasing/decreasing for $f$ using the sign of $f'$.
3. Determine concavity for $f$ by checking whether $f'$ is increasing or decreasing.
4. Combine them into a coherent shape. Increasing + concave up looks different from increasing + concave down.

A common error is to mix up what you’re sketching: if the given graph is $f'$, intercepts on that graph correspond to where $f$ has horizontal tangents—not where $f$ crosses the $x$-axis.

How to sketch $f'$ when you’re given $f$

If you’re given $f$ and asked about $f'$, you’re essentially graphing “slope as a function of $x$.” Conceptually:

• Where $f$ is increasing steeply, $f'$ is large and positive.
• Where $f$ is decreasing steeply, $f'$ is large in magnitude and negative.
• Where $f$ has a horizontal tangent, $f' = 0$.

If $f$ has a corner/cusp/vertical tangent, $f'$ may not exist there.

It helps to pick a few representative $x$-values, estimate the slope of $f$ at each, and plot those slope values on the $f'$ graph.

Putting the relationships in one place

The key ideas can be organized as “sign and trend.”

A note about “height” vs “slope” (avoiding a classic confusion)

Students often confuse these statements:

• “$f$ is positive” means the graph of $f$ is above the $x$-axis.
• “$f$ is increasing” means the graph goes up as you move right.

Similarly:

• “$f'$ is positive” does **not** mean $f$ is above the axis; it means $f$ is increasing.
• Zeros of $f'$ do **not** mean zeros of $f$.

Keeping “value” (height) separate from “rate of change” (slope) is the whole point of connecting these graphs.

Exam Focus

• Typical question patterns:
  • Given a graph of $f'$ (or a sign chart/table of $f'$), determine where $f$ is increasing/decreasing and identify local maxima/minima.
  • Given a graph of $f'$, determine where $f$ is concave up/down and locate inflection points by analyzing where $f'$ increases/decreases.
  • Sketch a possible graph of $f$ consistent with a provided graph of $f'$ (often with one point of $f$ given to anchor vertical position).
• Common mistakes:
  • Treating $f' = 0$ as automatically giving a local extremum; you must check a sign change in $f'$.
  • Confusing the zeros of $f'$ with the zeros of $f$.
  • Claiming an inflection point just because $f'' = 0$; you must confirm concavity changes (a sign change in $f''$, or equivalently a change in increasing/decreasing of $f'$).

Optimization Problems

An optimization problem asks you to find the maximum or minimum value of some quantity (area, cost, distance, volume, time, etc.) subject to given constraints. In calculus, the central idea is that maxima/minima often occur where a function’s derivative is zero or undefined—because at a “best possible” point, small changes usually don’t improve the outcome.

Optimization matters because it’s one of the most direct ways calculus models real decisions: minimizing materials in manufacturing, maximizing profit, designing containers, or finding the closest point to a location.

The big picture: objective function + constraints

Every optimization problem has two core ingredients:

1. Objective function: the quantity you want to optimize. Call it $Q$.
2. Constraints: equations/inequalities that limit possible choices.

Your job is usually to rewrite the objective function as a function of one variable, using the constraints. Then you can use derivatives to locate maxima/minima.

A very common place students go wrong is differentiating too early—before reducing to one variable. If you still have two variables, you either need more constraints or you haven’t used the given constraint fully.

Local vs absolute extrema (and why endpoints matter)

On AP problems, you’re often asked for an absolute maximum/minimum on a closed interval. The calculus fact you rely on is the Closed Interval Method:

• If $Q$ is continuous on $[a,b]$, then its absolute max and min occur either at:
  • endpoints $x=a$ or $x=b$, or
  • critical points inside $(a,b)$ where $Q' = 0$ or $Q'$ does not exist.

So optimization is not just “set derivative equal to zero.” You must also check endpoints if the domain is closed/bounded.

A standard step-by-step strategy (what you should actually do)

When you face an optimization prompt, a consistent workflow helps:

1. Draw a picture (even a rough one). Label variables.
2. Write the objective function $Q$ in terms of your variables.
3. Write the constraint equation(s) from the given conditions.
4. Reduce to one variable by solving the constraint for one variable and substituting.
5. Determine the domain of the remaining variable (physical constraints like lengths must be positive; also geometric limits).
6. Differentiate to find $Q'$.
7. Find critical points (solve $Q' = 0$ and note where $Q'$ undefined in the domain).
8. Decide max/min using:
   • the Closed Interval Method, or
   • the First Derivative Test, or
   • the Second Derivative Test (when appropriate).
9. Interpret your answer with units and a sentence (“maximum area occurs when …”).

Deciding whether you found a max or a min

After you find a critical point, you still must justify it produces the requested optimum.

First Derivative Test: If $Q'$ changes sign from positive to negative at a critical point, $Q$ has a local maximum there. If $Q'$ changes from negative to positive, $Q$ has a local minimum.

Second Derivative Test: If $Q'(c)=0$ and $Q''(c)>0$, then $Q$ has a local minimum at $x=c$. If $Q''(c)<0$, then $Q$ has a local maximum.

A common mistake is to apply the Second Derivative Test when $Q'(c)$ is not zero; it’s not valid in that case.

Example 1: Max area with fixed perimeter (classic fencing rectangle)

A farmer has $200$ meters of fencing to build a rectangular enclosure. What dimensions maximize the area?

1) Define variables. Let the rectangle have length $L$ and width $W$.

2) Objective function (area).

$A = LW$

3) Constraint (perimeter).

$2L + 2W = 200$

Solve for one variable, for example $W$:

$W = 100 - L$

4) Substitute into objective (one variable).

$A(L) = L(100 - L) = 100L - L^2$

5) Domain. Physical lengths require $L>0$ and $W>0$, so:

$0 < L < 100$

6) Differentiate.

$A'(L) = 100 - 2L$

7) Critical point. Set derivative to zero:

$100 - 2L = 0$

$L = 50$

Then $W = 100 - 50 = 50$.

8) Max or min? Since $A(L)$ is a downward-opening parabola (coefficient of $L^2$ is negative), the critical point is a maximum. Alternatively, check $A''(L)$:

$A''(L) = -2$

Because $A''(50) < 0$, it’s a local maximum; on this domain it’s also the absolute maximum.

Conclusion: The maximum area occurs when $L=W=50$ meters; the optimal rectangle is a square.

What to notice: Many geometry optimization problems end with symmetry (like a square), but you should still show it with calculus rather than guessing.

Example 2: Min distance from a point to a curve (closest point)

Find the point on the parabola $y = x^2$ closest to the point $P(0,3)$.

This is optimization because “closest” means “minimum distance.” The distance from a point $(x, x^2)$ on the parabola to $(0,3)$ is:

$D = \sqrt{(x-0)^2 + (x^2-3)^2}$

Minimizing $D$ is annoying because of the square root, but the square root is increasing, so minimizing $D$ is equivalent to minimizing $D^2$.

1) Objective function (squared distance).

$S(x) = D^2 = x^2 + (x^2 - 3)^2$

Expand:

$S(x) = x^2 + x^4 - 6x^2 + 9$

$S(x) = x^4 - 5x^2 + 9$

2) Differentiate.

$S'(x) = 4x^3 - 10x$

3) Critical points.

$4x^3 - 10x = 0$

Factor:

$2x(2x^2 - 5) = 0$

So:

$x = 0$

or

$2x^2 - 5 = 0$

$x^2 = \frac{5}{2}$

$x = \pm \sqrt{\frac{5}{2}}$

4) Evaluate the objective at candidates. Since the domain is all real numbers, we compare values of $S(x)$ at these points.

• $S(0) = 9$.
• For $x^2 = 5/2$, compute:

$S(x) = x^4 - 5x^2 + 9 = \left(\frac{5}{2}\right)^2 - 5\left(\frac{5}{2}\right) + 9$

$S(x) = \frac{25}{4} - \frac{25}{2} + 9$

$S(x) = \frac{25}{4} - \frac{50}{4} + \frac{36}{4}$

$S(x) = \frac{11}{4}$

Since $11/4 < 9$, the minimum occurs at $x = \pm \sqrt{5/2}$.

5) Give the actual points. On the parabola, $y=x^2$, so $y=5/2$.

Closest points:

$\left(\sqrt{\frac{5}{2}}, \frac{5}{2}\right)$

and

$\left(-\sqrt{\frac{5}{2}}, \frac{5}{2}\right)$

Common pitfall: Minimizing $D$ vs minimizing $D^2$. It’s fine to minimize $D^2$, but you must not forget at the end that the point is asked for, not just the distance.

Example 3: Minimizing material for a box (surface area with fixed volume)

A box with a square base has volume $V = 32$ cubic units. Find the dimensions that minimize surface area for an open-top box (no lid).

1) Variables. Let base side length be $x$ and height be $h$.

2) Constraint (volume).

$x^2 h = 32$

Solve for $h$:

$h = \frac{32}{x^2}$

3) Objective (surface area). For an open-top box:

• Base area: $x^2$
• Four sides: each is $xh$, so total side area is $4xh$

Thus:

$S = x^2 + 4xh$

Substitute $h$:

$S(x) = x^2 + 4x\left(\frac{32}{x^2}\right)$

Simplify:

$S(x) = x^2 + \frac{128}{x}$

4) Domain. $x>0$.

5) Differentiate and find critical points.

$S'(x) = 2x - \frac{128}{x^2}$

Set to zero:

$2x - \frac{128}{x^2} = 0$

$2x = \frac{128}{x^2}$

$2x^3 = 128$

$x^3 = 64$

$x = 4$

Then:

$h = \frac{32}{4^2} = 2$

6) Verify it’s a minimum. Use $S''(x)$:

$S''(x) = 2 + \frac{256}{x^3}$

For $x>0$, $S''(x) > 0$, so the critical point gives a minimum.

Conclusion: Minimum surface area occurs at base side $x=4$ and height $h=2$.

Why this makes sense: If the base is extremely small, you need a very tall box to keep volume fixed, increasing side area. If the base is huge, base area dominates. The minimum occurs at a balance point.

Optimization with contexts: what “reasonable domain” really means

AP problems often require you to state or use a domain. This isn’t just a formality—it prevents nonsense answers.

Examples of domain restrictions:

• Lengths/areas/volumes: variables must be positive.
• If you define $x$ as “distance from a corner along a side,” then $x$ must lie between $0$ and the side length.
• If you model revenue/cost with a demand function, you may need $x$ values that keep price nonnegative.

A subtle but common mistake is to find a critical point that is mathematically valid but physically impossible (like a negative radius). Always check units and constraints.

How curve sketching ideas support optimization

Optimization is basically curve sketching with a purpose. When you compute $Q'$ and analyze its sign, you’re doing the same work as “connecting $f$ and $f'$”:

• Sign of $Q'$ tells you where $Q$ increases/decreases.
• Where $Q'$ changes sign tells you maxima/minima.
• Sometimes $Q''$ helps you confirm the type of extremum via concavity.

Thinking this way makes optimization feel less like a “word problem trick” and more like an application of derivative behavior.

Exam Focus

• Typical question patterns:
  • “Find the dimensions that maximize area/minimize cost given a constraint” (perimeter, fixed volume, fixed surface area, fixed distance).
  • “Find the minimum distance” between a point and a curve (often by minimizing squared distance).
  • “Find absolute maximum/minimum on an interval” (explicitly requires checking endpoints plus critical points).
• Common mistakes:
  • Not reducing to one variable before differentiating, or using the constraint incorrectly when substituting.
  • Finding critical points but failing to test them (no endpoint check on a closed interval; no justification of max vs min).
  • Ignoring domain/physical feasibility, leading to impossible dimensions (negative lengths, values outside the geometric setup).