Unit 5: Analytical Applications of Differentiation

Extreme Value Theorem (EVT)

• The Extreme Value Theorem states that if a function is continuous on a closed interval \([a, b]\), it must have both an absolute maximum and an absolute minimum on that interval. This result guarantees that extrema exist, but it does not specify where they occur. The theorem is crucial for optimization problems because it justifies searching for maxima and minima within a given range.
• Continuity is essential; functions with breaks, jumps, or asymptotes may not have extrema in the expected locations. Similarly, the interval must be closed — open intervals may approach extreme values without ever attaining them. Recognizing these conditions prevents misapplication of the theorem.
• In practice, EVT guides the process of checking both endpoints and any critical points where \(f'(x) = 0\) or \(f'(x)\) is undefined. This ensures no potential extrema are overlooked. A common student mistake is forgetting to check endpoints, leading to missed maximum or minimum values.
• Graphically, EVT tells us that a continuous curve on a closed interval will have the highest and lowest points somewhere between or at the boundaries. This makes it easier to predict where extrema might occur before doing detailed calculations.
• EVT is especially useful in real-world contexts like production limits, profit maximization, or minimizing cost, where ensuring that maximum and minimum values exist is critical for making decisions.

Using the Candidates Test to Determine Absolute Extrema

• The candidates test is a systematic process for finding absolute extrema of a continuous function on a closed interval. The method involves evaluating the function at all critical points (where \(f'(x) = 0\) or undefined) and at the interval endpoints. This ensures no location of an extremum is missed.
• Step-by-step: (1) Differentiate the function to find \(f'(x)\). (2) Solve \(f'(x) = 0\) and check for undefined derivatives to find critical points. (3) Evaluate \(f(x)\) at each critical point and at \(x = a\) and \(x = b\). (4) The largest output is the absolute maximum, and the smallest output is the absolute minimum.
• Students often forget to check endpoints, leading to incomplete answers. Another common error is confusing relative extrema (local peaks or valleys) with absolute extrema, which requires comparing values across the entire domain in question.
• The candidates test is rooted in the Extreme Value Theorem — EVT assures us that maxima and minima exist, and the candidates test tells us exactly where they are. This connection reinforces why the theorem is taught first before introducing the test.
• In optimization problems, the candidates test provides a structured way to ensure that the chosen maximum or minimum is indeed the best possible solution within the given constraints.

Mean Value Theorem (MVT) and Rolle’s Theorem

• The Mean Value Theorem states that if a function is continuous on \([a, b]\) and differentiable on \((a, b)\), then there exists at least one \(c\) in \((a, b)\) such that \(f'(c) = \frac{f(b) - f(a)}{b - a}\). This means that at some point, the instantaneous rate of change (slope of the tangent) matches the average rate of change over the interval.
• Rolle’s Theorem is a special case of the MVT where \(f(a) = f(b)\). In this case, the theorem guarantees a point where \(f'(c) = 0\), meaning a horizontal tangent exists between \(a\) and \(b\). This is often used as a preliminary step in proofs involving MVT or curve behavior.
• Graphically, the MVT ensures the existence of a tangent line parallel to the secant line connecting \((a, f(a))\) and \((b, f(b))\). This is particularly important for analyzing motion, rates of change, and fairness arguments in applied settings.
• Both MVT and Rolle’s Theorem require continuity and differentiability. Students sometimes overlook these conditions, applying the theorems to functions with corners, cusps, or discontinuities, where they do not hold.
• These theorems are foundational for understanding how derivatives reflect the overall behavior of functions. They also serve as building blocks for proving deeper results in calculus, such as error bounds and the fundamental theorem of calculus.

Determining Intervals Over Which a Function is Increasing or Decreasing

• To determine where a function is increasing or decreasing, we examine the sign of its first derivative \( f'(x) \). If \( f'(x) > 0 \) on an interval, the function is increasing there, meaning its graph is rising as \( x \) moves from left to right. If \( f'(x) < 0 \), the function is decreasing, and its graph is falling.
• The process involves finding critical points (where \( f'(x) = 0 \) or undefined) and then testing the sign of \( f'(x) \) in each subinterval between them. This allows us to classify the behavior of the function piece by piece across its domain. Endpoints and discontinuities must be considered as boundaries for these intervals.
• Students often misinterpret zero derivatives as guaranteed extrema, but a flat slope can occur without being a maximum or minimum. Testing the intervals around the point ensures accurate classification. Similarly, failing to account for undefined derivatives can result in missing important changes in behavior.
• Graphically, positive slopes correspond to upward motion of the curve, while negative slopes correspond to downward motion. This relationship between derivative sign and graph shape is central to sketching and understanding functions without plotting every point.
• These intervals play a vital role in optimization problems, economic models, and physical applications like determining when an object speeds up or slows down. Without them, predicting trends and behavior would be guesswork rather than analysis.

Using the First Derivative Test to Determine Relative (Local) Extrema

• The First Derivative Test uses changes in the sign of \( f'(x) \) around critical points to determine whether they are relative maxima, minima, or neither. If \( f'(x) \) changes from positive to negative at a point, the function has a local maximum there. If it changes from negative to positive, the point is a local minimum.
• If there is no sign change in \( f'(x) \), the critical point is not an extremum — it could be an inflection point or a plateau. This makes the First Derivative Test more informative than simply finding where \( f'(x) = 0 \), as it distinguishes between peaks, valleys, and flat spots.
• The test involves: (1) finding critical points, (2) checking the sign of \( f'(x) \) in intervals around them, and (3) interpreting the results. Students often forget step 2, assuming that a zero derivative automatically means an extremum, which is incorrect.
• In motion problems, a local maximum might represent the highest point reached by an object, while a local minimum could represent the lowest position or slowest speed. The test thus connects directly to physical applications where turning points are important.
• The First Derivative Test also complements the Second Derivative Test, which uses concavity to confirm extrema. Using both provides stronger evidence and can save time when one method is inconclusive.

Derivatives and Properties of Functions

• Derivatives reveal important properties of functions, including monotonicity (increasing or decreasing behavior), concavity, and the presence of inflection points. The first derivative \( f'(x) \) indicates slope behavior, while the second derivative \( f''(x) \) provides insight into how the slope itself is changing.
• For example, \( f'(x) > 0 \) means the function is increasing, while \( f'(x) < 0 \) means it is decreasing. Similarly, \( f''(x) > 0 \) indicates concave up behavior (like a cup), and \( f''(x) < 0 \) indicates concave down behavior (like a frown). These conditions are central to accurate graph sketching.
• Derivatives also help in identifying points of inflection, where concavity changes. This occurs when \( f''(x) = 0 \) or undefined, but must be verified by checking concavity on both sides of the point to confirm the change.
• Properties such as symmetry can be connected to derivatives as well — for example, odd functions have certain derivative patterns, and even functions have others. Understanding these relationships can make function analysis more efficient.
• In real-world scenarios, derivative properties can model velocity and acceleration in physics, marginal cost and revenue in economics, and rates of change in biology and chemistry. Recognizing these connections ensures a deeper understanding of calculus beyond abstract computation.

Determining Concavity of Functions Over Their Domains

• Concavity describes how the slope of a function is changing, which we determine using the sign of the second derivative \( f''(x) \). If \( f''(x) > 0 \) on an interval, the graph is concave up, meaning slopes are increasing and the curve opens upward. If \( f''(x) < 0 \), the graph is concave down, meaning slopes are decreasing and the curve opens downward.
• Finding concavity involves computing \( f''(x) \), identifying points where \( f''(x) = 0 \) or is undefined, and testing intervals between these points. These boundary points are candidates for inflection points where concavity changes.
• It’s important to note that \( f''(x) = 0 \) alone does not guarantee an inflection point — the concavity must actually change from positive to negative or vice versa. This verification step is often overlooked, leading to false conclusions.
• Concavity analysis also helps anticipate the overall shape of the graph and is used in conjunction with first derivative tests for more complete graph sketching. It can provide early insight into the “feel” of the function without plotting numerous points.
• In applied contexts, concavity can indicate acceleration trends: concave up corresponds to increasing velocity, and concave down corresponds to decreasing velocity. This is especially useful in physics and engineering applications.

Using the Second Derivative Test to Determine Extrema

• The Second Derivative Test provides a quicker way to determine whether a critical point is a local maximum or minimum. If \( f'(c) = 0 \) and \( f''(c) > 0 \), the function has a local minimum at \( x = c \) because the graph is concave up. If \( f'(c) = 0 \) and \( f''(c) < 0 \), the function has a local maximum because the graph is concave down.
• If \( f''(c) = 0 \) or is undefined, the test is inconclusive, and we must revert to the First Derivative Test or another method to classify the point. This prevents misclassification when the curvature is flat or changing unpredictably at the critical point.
• While the Second Derivative Test is faster than testing sign changes in the first derivative, it only works when \( f'(x) \) is already zero and the second derivative has a clear sign. It cannot be applied in isolation without this prerequisite.
• Graphically, this test aligns with the intuitive idea that a “cup” shape at a critical point (concave up) indicates a minimum and a “cap” shape (concave down) indicates a maximum. This visualization reinforces the calculus-based classification.
• In real-world problems, this test can save significant time when optimizing production, finding turning points in motion problems, or analyzing profit functions where derivative computations are already in progress.

Sketching Graphs of Functions and Their Derivatives

• Sketching a graph using derivatives involves combining information about intercepts, asymptotes, increasing/decreasing intervals, relative extrema, concavity, and inflection points. The first derivative tells us where slopes are positive or negative, and the second derivative tells us about curvature.
• The process typically starts with domain analysis, then moves to finding critical points and testing for relative extrema. Next, concavity is determined, and finally, all information is combined to produce an accurate and coherent graph.
• It’s essential to understand how the graph of \( f'(x) \) relates to \( f(x) \) — zeros of \( f'(x) \) correspond to horizontal tangents of \( f(x) \), and positive/negative regions of \( f'(x) \) indicate increasing/decreasing behavior of \( f(x) \). Similarly, zeros of \( f''(x) \) indicate possible inflection points in \( f(x) \).
• Students often struggle by focusing only on point-plotting rather than interpreting derivative behavior, which leads to missed features like asymptotic behavior or curvature changes. A full derivative analysis ensures a graph matches the function’s true shape.

How to Sketch Graphs of Functions and Their Derivatives

• Start by identifying the domain of the function \( f(x) \). Note any restrictions such as division by zero or square roots of negative numbers. These restrictions define the x-values you can include in your sketch and may indicate vertical asymptotes or gaps in the graph.
• Find the first derivative \( f'(x) \) to determine intervals of increase and decrease. Solve \( f'(x) = 0 \) to find critical points, then use a sign chart for \( f'(x) \) to see where the slope is positive (increasing) and negative (decreasing). This will outline the overall shape and direction changes.
• Use the second derivative \( f''(x) \) to determine concavity. Solve \( f''(x) = 0 \) to find possible inflection points, and check the sign of \( f''(x) \) on each interval to determine concave up (positive) and concave down (negative) regions. This adds curvature detail to your sketch.
• Plot key points: x- and y-intercepts, critical points, and inflection points. Label whether each critical point is a local maximum, local minimum, or saddle point based on derivative tests. Include asymptotes and any end behavior using limits or the degree of polynomials/rational functions.
• Once \( f(x) \) is sketched, you can sketch \( f'(x) \) by remembering: \( f'(x) = 0 \) where \( f(x) \) has horizontal tangents, \( f'(x) > 0 \) where \( f(x) \) is increasing, and \( f'(x) < 0 \) where \( f(x) \) is decreasing. Similarly, sketch \( f''(x) \) based on the slope of \( f'(x) \). Always match changes in sign between derivatives and their original function to maintain accuracy.

Connecting a Function, Its First Derivative, and Its Second Derivative

• The first derivative \( f'(x) \) describes the slope of the original function \( f(x) \), and the second derivative \( f''(x) \) describes the rate of change of that slope. Together, they reveal both the direction and curvature of the graph.
• When \( f'(x) > 0 \), the function is increasing; when \( f'(x) < 0 \), the function is decreasing. Similarly, \( f''(x) > 0 \) means the graph is concave up, and \( f''(x) < 0 \) means the graph is concave down.
• Critical points from \( f'(x) = 0 \) can be classified as maxima, minima, or inflection points by considering \( f''(x) \). Inflection points occur when \( f''(x) \) changes sign, even if \( f'(x) \) is not zero.
• In applied settings, this combined analysis can model motion: \( s(t) \) represents position, \( s'(t) \) velocity, and \( s''(t) \) acceleration. This shows how calculus seamlessly connects geometric interpretation with physical meaning.
• Mastering this connection is essential for graph sketching, optimization, and solving applied problems, as it allows prediction of shape and turning points without full plotting.

Optimization Problems

• Optimization problems involve finding the maximum or minimum value of a function under given constraints. They often appear in contexts such as maximizing area, minimizing cost, or optimizing efficiency in real-world scenarios.
• The process includes defining the variable(s), writing the objective function, applying constraints to reduce variables, finding the derivative, and solving for critical points. Endpoints must also be checked if applicable.
• Using the First or Second Derivative Test confirms whether a critical point is a maximum or minimum. Without this step, one risks misidentifying an extremum, especially in non-symmetric or complex functions.
• Optimization is used across disciplines — in physics for minimizing travel time, in business for maximizing profit, and in engineering for material efficiency. The same calculus process applies despite context changes.
• Clear interpretation of the result with correct units is crucial; even a correct numerical answer is incomplete without explaining what it means in context.

Optimization Problems in Real Contexts

Concept and Problem-Solving Process

• Optimization problems involve finding the maximum or minimum value of a quantity under given constraints, and they are common in physics, engineering, economics, and everyday decision-making. In AP Calculus, these problems typically require translating a real-world scenario into a mathematical model, then using derivatives to locate and classify extrema. The goal is to determine the most efficient, cost-effective, or practical solution based on the problem’s context.
• The process begins by defining the objective function, which represents the quantity to be maximized or minimized. Constraints from the problem are used to express the objective function in terms of a single variable, often requiring geometric relationships, trigonometry, or physical formulas. Properly setting up this relationship is the foundation of a correct and efficient solution.
• Once the objective function is established, its derivative is calculated to find critical points. Critical points occur where the derivative equals zero or does not exist, and these represent potential maxima or minima. Testing each critical point, as well as endpoints of the domain when applicable, determines the optimal value.
• The second derivative test or sign analysis of the first derivative can confirm whether a critical point is a maximum or minimum. In real contexts, this verification step ensures the solution makes practical sense—such as confirming that a container’s dimensions truly give the largest possible volume rather than a smallest possible one.
• Examples of optimization problems include designing a cylindrical can with the least surface area for a given volume, maximizing revenue by adjusting product price, or determining the fastest route that minimizes travel time. Clear labeling of variables, correct use of units, and interpretation of results in context are essential for full credit on AP Calculus optimization questions.

Connecting Related Rates and Optimization in Real Contexts

• Some real-world problems combine related rates and optimization, requiring both the rate of change of quantities and the identification of optimal values under constraints. This is common in engineering and physics problems involving dynamic systems.
• The key is to first set up the related rates equation by differentiating all relevant relationships with respect to time. Then, identify an expression to maximize or minimize, applying the derivative tests for optimization.
• These problems often require interpreting physical meaning — for example, adjusting a filling rate to maximize volume change at a specific moment or minimizing the rate of energy loss in a system.
• They reinforce the importance of unit consistency and real-world feasibility. A mathematically correct solution might be impractical if constraints or limits are ignored.
• Solving them builds skill in connecting multiple calculus concepts at once, which is essential for tackling complex AP Free Response Questions.

Behaviors of Implicit Relations

• Implicit relations describe functions indirectly, without being solved for one variable in terms of another. Differentiating these requires implicit differentiation, which applies the chain rule to both sides of the equation.
• Critical points and slope behavior for implicit curves are determined by differentiating and then solving for \( \frac{dy}{dx} \). This allows analysis of curves like circles, ellipses, and more complex algebraic curves that aren’t easily written in explicit form.
• The derivative still provides all the information about increasing/decreasing behavior, concavity, and extrema, even if the function is not isolated. These properties can be analyzed locally around given points.
• Graphical behavior such as symmetry and multiple-valued outputs can be studied by looking at how \( \frac{dy}{dx} \) behaves across different portions of the curve.
• Implicit differentiation is also necessary in related rates problems involving geometric constraints and in higher-level applications like polar coordinates and conic sections.

Example Problem 1:

Problem: Sketch the graph of \( f(x) = x^3 - 3x^2 - 9x + 5 \) and its first derivative. Identify intervals of increase/decrease, local maxima and minima, concavity, and inflection points.

Solution:

• Step 1- Domain: This is a cubic polynomial, so the domain is all real numbers. There are no restrictions.
• Step 2- First derivative: \( f'(x) = 3x^2 - 6x - 9 = 3(x^2 - 2x - 3) = 3(x - 3)(x + 1) \). Critical points are \( x = -1 \) and \( x = 3 \).
• Step 3- Sign chart for \( f'(x) \):
  • For \( x < -1 \), pick \( x = -2 \): \( f'(-2) = 3(4 + 4 - 3) > 0 \) → increasing.
  • For \( -1 < x < 3 \), pick \( x = 0 \): \( f'(0) = -9 < 0 \) → decreasing.
  • For \( x > 3 \), pick \( x = 4 \): \( f'(4) = 3(16 - 8 - 3) > 0 \) → increasing.
  Thus, local max at \( x = -1 \), local min at \( x = 3 \).
• Step 4- Second derivative: \( f''(x) = 6x - 6 = 6(x - 1) \). Inflection point at \( x = 1 \).
• Step 5- Concavity: For \( x < 1 \), \( f''(x) < 0 \) → concave down. For \( x > 1 \), \( f''(x) > 0 \) → concave up.
• Step 6- Key points:
  • Local max: \( f(-1) = 10 \)
  • Local min: \( f(3) = -22 \)
  • Inflection: \( f(1) = -6 \)
• Step 7- Sketch: Plot the intercepts, maxima, minima, and inflection point. Draw increasing/decreasing and concavity behavior. For \( f'(x) \), plot a parabola with zeros at \( -1 \) and \( 3 \) and vertex at \( x = 1 \).

Example Problem 2:

Problem: A rectangular sheet of cardboard measures \(20\text{ in} \times 30\text{ in}\). Squares of side length \(x\) are cut from each corner, and the sides are folded up to form an open-top box. Determine the value of \(x\) that maximizes the box’s volume, and report the maximum volume. Explain each step clearly so the method is reusable on similar problems.

Solution:

• Let \(x\) be the cutout size (in inches). After cutting and folding, the base dimensions become \( (20 - 2x) \) by \( (30 - 2x) \), and the height is \(x\). Therefore, the volume as a function of \(x\) is \( V(x) = x(20 - 2x)(30 - 2x) \). We will optimize \(V(x)\) over physically meaningful values of \(x\).
• First simplify the volume function for differentiation. Multiply \( (20 - 2x)(30 - 2x) = 600 - 100x + 4x^2 \), so \( V(x) = x(600 - 100x + 4x^2) = 600x - 100x^2 + 4x^3 \). The domain is constrained by positive dimensions: \(20 - 2x > 0 \Rightarrow x < 10\) and \(30 - 2x > 0 \Rightarrow x < 15\), with \(x > 0\). Hence, the feasible domain is \( 0 < x < 10 \).
• Differentiate to find critical points: \( V'(x) = 600 - 200x + 12x^2 \). Set \( V'(x)=0 \Rightarrow 12x^2 - 200x + 600 = 0 \), divide by 4 to get \( 3x^2 - 50x + 150 = 0 \). Solve the quadratic: \( x = \dfrac{50 \pm \sqrt{50^2 - 4\cdot 3\cdot 150}}{2\cdot 3} = \dfrac{50 \pm \sqrt{700}}{6} = \dfrac{50 \pm 10\sqrt{7}}{6} \). Only \( x = \dfrac{50 - 10\sqrt{7}}{6} \approx 3.924 \) lies in \( (0,10) \); the other root is about \(12.743\) and is not feasible.
• Classify the critical point with the second derivative or candidates test. Compute \( V''(x) = -200 + 24x \), so \( V''(3.924) \approx -200 + 94.2 \approx -105.8 < 0 \), indicating a local maximum. Endpoints \(x \to 0^+\) and \(x=10\) both yield \(V=0\), so the interior critical point gives the absolute maximum on the domain. This verifies the choice is optimal for the physical box.
• Report the optimal dimensions and maximum volume. At \( x^* \approx 3.924\text{ in} \), the base is \( 20 - 2x^* \approx 12.15\text{ in} \) by \( 30 - 2x^* \approx 22.15\text{ in} \), and the height is \( \approx 3.92\text{ in} \). The maximum volume is \( V(x^*) = 600x^* - 100(x^*)^2 + 4(x^*)^3 \approx 1056 \text{ in}^3 \) (to the nearest cubic inch). Always include units and a brief check that dimensions remain positive and practical.

Common Misconceptions

• Students often forget to identify the physical or logical domain before solving. For example, in a box problem, \(x\) must be positive and small enough so the base dimensions remain positive. Solving without these restrictions can lead to “optimal” answers that are impossible to build in real life.
• Another common error is neglecting to test endpoints when using the candidates test. Some optimization problems achieve maxima or minima at the boundary of the domain, so ignoring endpoints can cause the true optimum to be missed entirely.
• Many students fail to interpret the meaning of variables in context. In related rates or geometric optimization, it’s easy to treat \(x\) as just an algebraic symbol without realizing it represents a real measurement (length, radius, time). This disconnect can cause nonsensical answers with incorrect units.
• Some learners rely solely on the first derivative test but misinterpret sign charts, especially if \(f'(x)\) changes sign more than once or has undefined points. Careful labeling of intervals and signs is essential to avoid concluding that a saddle point is an extremum.
• It is also common to stop after finding the critical number without confirming whether it yields a maximum or minimum. The second derivative test or a candidates test should always be applied to ensure that the solution matches the problem’s goal (maximization or minimization).