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Calculus - Applications of the Derivative

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    Study Guide Calculus – Applications of the Derivative 1. Critical Points When studying functions and their graphs, critical points are very important. These points often help us understand where a function reaches a maximum , minimum , or changes its overall behavior. What Is a Critical Point? A point ((x, f(x))) is called a critical point of a function (f(x)) if: • (x) is in the domain of the function, and • either o the derivative at that point is zero, (f'(x) = 0), or o the derivative does not exist at that point. Geometric Meaning At a critical point, one of the following happens to the tangent line of the graph: • It is horizontal (slope = 0), • It is vertical , or • It does not exist . These points are key places where the graph may change direction or shape. Example 1: Finding Critical Points of a Polynomial Find all critical points of Step 1: Identify the Domain Since (f(x)) is a polynomial , its domain is all real numbers .

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    Study Guide Step 2: Find the Derivative Step 3: Set the Derivative Equal to Zero Factor: Solve: Step 4: Find the Corresponding Function Values Final Answer The critical points are: Example 2: Critical Points of a Trigonometric Function Find all critical points of

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    Study Guide Step 1: Understand the Domain The domain is restricted to the closed interval ([0, 2 π ). Step 2: Find the Derivative Step 3: Set the Derivative Equal to Zero This happens at: Step 4: Find the Function Values Final Answer The critical points are:

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    Study Guide Key Takeaways To find critical points: 1. Find the derivative of the function. 2. Set the derivative equal to zero (and also check where it does not exist). 3. Make sure the points are in the domain . 4. Evaluate the function at those (x) - values. Critical points are powerful tools that help us understand how a function behaves — and they show up everywhere in calculus! 2. Extreme Value Theorem Critical points are especially useful when we want to find the maximum and minimum values of a function on a specific interval. This is where the Extreme Value Theorem (EVT) comes in. What Does the Extreme Value Theorem Say? The Extreme Value Theorem guarantees that a function will reach both a highest value (maximum) and a lowest value (minimum) under the right conditions. The theorem states: • If a function (f(x)) is continuous on a closed interval ([a, b]), then (f(x)) has both a maximum and a minimum value on that interval. In short: • Continuous function • Closed interval (includes endpoints) → A maximum and minimum must exist .

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    Study Guide How to Use the Extreme Value Theorem To apply the theorem, follow these steps: 1. Check continuity Make sure the function is continuous on the given closed interval. 2. Find all critical points Compute the derivative and find where: o (f'(x) = 0), or o (f'(x)) does not exist (Only include points that lie inside the interval .) 3. Evaluate the function Find the function value at: o each critical point in the interval, and o both endpoints of the interval. 4. Compare values o The largest value is the maximum. o The smallest value is the minimum. Example 1: Trigonometric Function Find the maximum and minimum values of Step 1: Check Continuity Sine and cosine are continuous, so (f(x)) is continuous on ([0, 2 π ]).

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    Study Guide Step 2: Identify Critical Points From earlier work, the critical points are: Step 3: Check the Endpoints Step 4: Compare All Values • Maximum value: ( √2 ) at (x = π/4 ) • Minimum value: ( - √2 ) at (x = 5 π/4 ) Final Answer • Maximum: ( √2 ) • Minimum: ( - √2 ) Note: In this example, both the maximum and minimum occur at critical points . Example 2: Polynomial Function Find the maximum and minimum values of Step 1: Check Continuity Polynomials are continuous everywhere, so the function is continuous on ([ - 2, 2]).

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    Study Guide Step 2: Find the Derivative Set the derivative equal to zero: Solve: Step 3: Keep Only Points in the Interval • (x = 9/4 ) is not in ([ - 2, 2]), so we ignore it. • The only critical point in the interval is (x = 0). Step 4: Evaluate the Endpoints Step 5: Compare All Values • Maximum value: 39 at (x = - 2) • Minimum value: – 9 at (x = 2) Final Answer • Maximum: 39 • Minimum: – 9

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    Study Guide Important Note: This example shows why the closed interval matters . Even though there was a critical point, the maximum and minimum occurred at the endpoints , not at the critical point. Key Takeaways When finding absolute maximums and minimums: • Always check critical points AND endpoints • Only consider values inside the given interval • The Extreme Value Theorem guarantees the answer exists — but you still have to find it 3. Mean Value Theorem The Mean Value Theorem (MVT) connects two important ideas in calculus: • the slope of a secant line (average rate of change), and • the slope of a tangent line (instantaneous rate of change). In simple terms, the theorem says that somewhere between two points, the instantaneous slope matches the average slope . Figure 1 The Mean Value Theorem.

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    Study Guide Statement of the Mean Value Theorem If a function (f(x)) satisfies both of the following conditions: • it is continuous on the closed interval ([a, b]), and • it is differentiable on the open interval ((a, b)), then there exists at least one number (c) in ((a, b)) such that What Does This Mean Geometrically? • The fraction represents the slope of the secant line joining the points ((a, f(a))) and ((b, f(b))). • The value (f'(c)) represents the slope of the tangent line at some point (x = c). The Mean Value Theorem guarantees that at least one tangent line between (a) and (b) is parallel to the secant line . A Special Case to Remember If (f(a) = f(b)), then the slope of the secant line is zero. In this case, the theorem guarantees at least one critical point where This idea is closely related to Rolle’s Theorem . Example 1 : Verifying the Mean Value Theorem Verify the conclusion of the Mean Value Theorem for

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    Study Guide Step 1: Check the Conditions • (f(x)) is a polynomial → continuous everywhere • Polynomials are also differentiable everywhere So, the Mean Value Theorem applies on ([ - 2, 3]). Step 2: Find the Slope of the Secant Line Evaluate the function: Now compute the slope: So, the average rate of change is – 2 . Step 3: Find the Derivative Step 4: Set the Derivative Equal to the Secant Slope

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