Unit 5: Analytical Applications of Differentiation

Mean Value Theorem (MVT)

Definition and Conditions

• The Mean Value Theorem states that if a function \( f(x) \) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one number \( c \in (a, b) \) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \). This slope represents the average rate of change over the interval.
• Continuity ensures there are no breaks or jumps in the graph, while differentiability ensures there are no sharp corners or vertical tangents. Without these, MVT does not apply.
• The slope of the secant line between \( a \) and \( b \) equals the slope of the tangent line at \( c \). This means the graph has at least one point where its instantaneous rate of change matches its average rate of change.
• MVT is an extension of Rolle’s Theorem, which is the special case where \( f(a) = f(b) \) and thus \( f'(c) = 0 \). If \( f(a) \neq f(b) \), the tangent slope matches the secant slope instead of zero.
• In applications, MVT is used to prove that certain speeds or rates must have occurred during an interval, often in motion, population growth, or economics problems.

Example Problem

Suppose a car travels 120 km in 2 hours. The average speed is \( \frac{120}{2} = 60 \) km/h. If the car’s speed is continuous and differentiable over the trip, MVT guarantees there was at least one moment when the car’s instantaneous speed was exactly 60 km/h.

Extreme Value Theorem (EVT)

Definition and Conditions

• The Extreme Value Theorem states that if a function \( f(x) \) is continuous on a closed interval \([a, b]\), then \( f(x) \) must attain both an absolute maximum and an absolute minimum value on that interval. These extrema may occur at critical points or at the endpoints.
• Continuity is essential; without it, the function could have holes or jumps that prevent certain extreme values from existing. The requirement of a closed interval ensures the range is bounded and includes its endpoints.
• The process to find extrema involves finding \( f'(x) \), setting it to zero to find critical points, and then evaluating \( f(x) \) at these points and at the endpoints. The largest value is the absolute maximum, and the smallest is the absolute minimum.
• EVT does not require differentiability; corners and cusps are allowed as long as the function is continuous. However, differentiability is required for certain optimization methods.
• EVT is often applied in real-world contexts such as finding maximum profit, minimum cost, or highest elevation within a given range of inputs.

Example Problem

Find the absolute extrema of \( f(x) = x^2 - 4x + 1 \) on \([0, 5]\).

Solution:

• Step 1: Differentiate: \( f'(x) = 2x - 4 \).
• Step 2: Set \( f'(x) = 0 \) → \( 2x - 4 = 0 \) → \( x = 2 \) (critical point).
• Step 3: Evaluate \( f(x) \) at \( x = 0, 2, 5 \):
  • \( f(0) = 0^2 - 4(0) + 1 = 1 \)
  • \( f(2) = (2)^2 - 4(2) + 1 = -3 \)
  • \( f(5) = (5)^2 - 4(5) + 1 = 6 \)
• Step 4: The absolute minimum is \( -3 \) at \( x = 2 \), and the absolute maximum is \( 6 \) at \( x = 5 \).

Finding Absolute Extrema

Definition and Concept

• Absolute extrema are the highest (absolute maximum) or lowest (absolute minimum) values of a function on a given interval, considering both interior points and endpoints. These points represent the extreme vertical values the function can reach over the specified domain.
• Unlike relative extrema, which only compare a point to nearby values, absolute extrema compare a point to every other point on the interval. They are the “global” highest and lowest points, not just local peaks or valleys.
• The Extreme Value Theorem guarantees that a continuous function on a closed interval \([a,b]\) has both an absolute maximum and an absolute minimum somewhere on the interval.
• Absolute extrema can occur at critical points inside the interval (where \( f'(x) = 0 \) or \( f'(x) \) is undefined) or at the endpoints \(x = a\) and \(x = b\).
• Finding absolute extrema is essential in real-world applications where you need to optimize over a fixed range, such as determining the maximum profit, highest stress on a beam, or lowest temperature in a 24-hour cycle.

Procedure for Finding Absolute Extrema

• Ensure the function is continuous on the given interval. If it’s not, the Extreme Value Theorem does not apply and extrema may not exist.
• Find all critical points in the open interval \((a,b)\) by solving \( f'(x) = 0 \) or identifying where \( f'(x) \) is undefined but \( f(x) \) is defined.
• Evaluate the original function \( f(x) \) at each critical point and at both endpoints \(a\) and \(b\).
• Compare all obtained function values. The largest value is the absolute maximum, and the smallest value is the absolute minimum.
• Always check endpoints — many absolute extrema occur there, especially when the slope does not change sign within the interval.

Common Mistakes

• Forgetting to check endpoints, leading to missing the true absolute extrema.
• Using \( f'(x) \) values instead of \( f(x) \) values when comparing heights of points.
• Assuming the existence of extrema without checking continuity; a discontinuity may prevent the Extreme Value Theorem from applying.
• Only finding relative extrema and thinking they are absolute without verifying against all points in the interval.
• Not considering intervals that include vertical asymptotes or undefined regions where the function may approach infinity.

Rolle’s Theorem and Connections to the Mean Value Theorem (MVT)

• Rolle’s Theorem states that if a function \( f(x) \) is continuous on the closed interval \([a,b]\), differentiable on the open interval \((a,b)\), and \( f(a) = f(b) \), then there exists at least one \( c \in (a,b) \) such that \( f'(c) = 0 \). This guarantees a horizontal tangent within the interval.
• To apply Rolle’s Theorem, you must first verify all three conditions: continuity on \([a,b]\), differentiability on \((a,b)\), and equal function values at the endpoints. Skipping any check may lead to incorrect assumptions.
• The Mean Value Theorem (MVT) generalizes Rolle’s Theorem by removing the requirement that \( f(a) = f(b) \). MVT states that if \( f \) is continuous on \([a,b]\) and differentiable on \((a,b)\), there exists a point \( c \) such that \( f'(c) = \frac{f(b)-f(a)}{b-a} \).
• In applied contexts, MVT tells us that there is some instant where the instantaneous rate of change equals the average rate of change over the interval. For example, in motion problems, there’s a time where instantaneous velocity equals average velocity.
• Rolle’s Theorem is essentially a special case of MVT when the average rate of change is zero. Graphically, it means there’s a flat tangent line somewhere in the interval if the function starts and ends at the same height.

First Derivative Test for Relative Extrema

• The First Derivative Test uses sign changes in \( f'(x) \) around critical points to determine whether those points are relative minima, maxima, or neither. Critical points occur where \( f'(x) = 0 \) or \( f'(x) \) is undefined, provided \( f(x) \) is continuous there.
• If \( f'(x) \) changes from positive to negative at a critical point \( c \), \( f(c) \) is a relative maximum. If \( f'(x) \) changes from negative to positive, \( f(c) \) is a relative minimum.
• If \( f'(x) \) does not change sign across \( c \), the critical point is not a relative extremum — it could be a point of inflection or a plateau.
• This test is especially useful for functions where the second derivative is complex or undefined, making the Second Derivative Test difficult to apply.
• When sketching graphs, the First Derivative Test provides directional behavior before and after critical points, helping identify peaks and valleys quickly without computing second derivatives.

Second Derivative Test for Concavity and Inflection Points

• The Second Derivative Test evaluates \( f''(x) \) at critical points to classify them: if \( f''(c) > 0 \), \( f(c) \) is a local minimum; if \( f''(c) < 0 \), \( f(c) \) is a local maximum. If \( f''(c) = 0 \), the test is inconclusive.
• Concavity describes how a curve bends: \( f''(x) > 0 \) means the function is concave up (cup-shaped), and \( f''(x) < 0 \) means concave down (cap-shaped).
• Inflection points occur where the concavity changes from up to down or vice versa. At these points, \( f''(x) \) is typically zero or undefined, but a sign change in \( f''(x) \) is necessary to confirm an actual change in concavity.
• The Second Derivative Test is quicker than the First Derivative Test if the function’s second derivative is easy to compute and evaluate.
• In applications, concavity reveals acceleration trends in motion problems, curvature in design and engineering, and cost/revenue acceleration in economics, making inflection points critical for optimization.

Sketching Graphs Using First and Second Derivatives

• First derivatives \( f'(x) \) indicate where a function is increasing (\( f'(x) > 0 \)) or decreasing (\( f'(x) < 0 \)). These intervals help identify local maxima and minima when combined with critical points. This is crucial for determining turning points and general graph structure.
• Second derivatives \( f''(x) \) reveal concavity: concave up when \( f''(x) > 0 \) and concave down when \( f''(x) < 0 \). Points where \( f''(x) = 0 \) and concavity changes are inflection points.
• To sketch a graph: first find critical points from \( f'(x) = 0 \) or undefined, then test intervals for increasing/decreasing behavior. Use \( f''(x) \) to determine concavity and locate inflection points.
• End behavior is determined from limits as \( x \to \pm \infty \), often influenced by leading terms in polynomials or asymptotic behavior in rational/exponential functions.
• By combining increasing/decreasing intervals, concavity, asymptotes, and intercepts, you can create an accurate representation of a function without plotting many points.

Optimization Problems (Applied Max/Min)

• Optimization problems require finding the maximum or minimum value of a quantity under given constraints. They are common in physics, economics, and engineering.
• The process involves defining a function for the quantity to be optimized, expressing it in terms of a single variable (using constraints), and finding its critical points by solving \( f'(x) = 0 \) or undefined.
• Check both critical points and endpoints (if applicable) to determine absolute extrema. Use first or second derivative tests to verify whether a point is a maximum or minimum.
• Units must be consistent, and constraints are often geometric (e.g., perimeter, area, volume) or cost/efficiency-based in real-world applications.
• Clear labeling of variables and translating word problems into equations is essential for solving optimization problems accurately.

Connecting Derivatives to Shape of a Graph

• The first derivative \( f'(x) \) controls the slope of the tangent line at any point on the graph. Positive slopes create upward trends, negative slopes create downward trends.
• The second derivative \( f''(x) \) indicates the acceleration of change in the slope, giving insight into the “bending” of the graph.
• Changes from positive to negative in \( f'(x) \) indicate local maxima; changes from negative to positive indicate local minima.
• Concave up shapes (\( f''(x) > 0 \)) open like a cup and are associated with minima, while concave down shapes (\( f''(x) < 0 \)) open like a cap and are associated with maxima.
• Understanding both derivatives together allows you to describe the entire behavior of a graph without relying solely on raw plotting — this is critical in modeling and predicting real-world trends.

Analyzing Curves Defined by Equations (Implicit Relations)

• Some curves are defined by equations that cannot be easily written in explicit form \( y = f(x) \), such as circles \( x^2 + y^2 = r^2 \) or ellipses. In these cases, implicit differentiation is used to find derivatives without first isolating \( y \).
• To differentiate implicitly, take the derivative of both sides of the equation with respect to \( x \), applying the chain rule to any \( y \)-terms since \( y \) is a function of \( x \). This introduces \( \frac{dy}{dx} \) into the equation, which can then be solved for explicitly.
• Implicit differentiation is particularly important for relations that represent physical constraints or geometric shapes where solving for \( y \) is cumbersome or impossible.
• Once \( \frac{dy}{dx} \) is found, it can be used to analyze slopes of tangents, identify horizontal or vertical tangents, and study the shape of the curve.
• Applications include finding slopes of tangent lines in conic sections, related rates involving circular motion, and analyzing special curves like cardioids or lemniscates.

Analyzing Curves Defined by Equations (Implicit Relations) – How To

• Step 1: Identify that the curve is not written explicitly as \( y = f(x) \) but is instead given by an equation involving both \( x \) and \( y \), such as \( x^2 + xy + y^2 = 7 \).
• Step 2: Differentiate both sides of the equation with respect to \( x \), treating \( y \) as a function of \( x \). Apply the chain rule when differentiating terms with \( y \) (e.g., \( \frac{d}{dx}[y^2] = 2y \frac{dy}{dx} \)).
• Step 3: Collect all terms involving \( \frac{dy}{dx} \) on one side of the equation, and all other terms on the opposite side.
• Step 4: Factor out \( \frac{dy}{dx} \) and solve algebraically for it to find the slope of the tangent line at any point on the curve.
• Step 5: To find the slope at a specific point, substitute both \( x \) and \( y \) values into the \( \frac{dy}{dx} \) formula you found.

Applying Derivatives to Problems Involving Rates of Change in Other Contexts (Multi-Variable Situations) – How To

• Step 1: For problems where multiple quantities depend on each other, make sure all variables are linked through one main equation before differentiating.
• Step 2: Differentiate implicitly if necessary, especially when the variables are not isolated (common in physics and engineering problems).
• Step 3: Always check dimensional consistency — this helps avoid plugging in mismatched units for rates (e.g., meters per second vs. feet per second).
• Step 4: In applied contexts, interpret the sign and magnitude of your result in terms of the real-world situation (e.g., speed increasing, tank draining).
• Step 5: Clearly state the final answer with correct units and a conclusion sentence describing what the result means physically.