Calculus - Applications of the Derivative - Quick Guides

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Study GuideCalculus–Applications of the Derivative1.Critical PointsWhen studying functions and their graphs,critical pointsare very important. These points often helpus understand where a function reaches amaximum,minimum, or changes its overall behavior.What Is a Critical Point?A point ((x, f(x))) is called acritical pointof a function (f(x)) if:•(x) is in thedomainof the function,and•eitherothe derivative at that point is zero, (f'(x) = 0),orothe derivative doesnot existat that point.Geometric MeaningAt a critical point, one of the following happens to the tangent line of the graph:•It ishorizontal(slope = 0),•It isvertical, or•Itdoes not exist.These points are key places where the graph may change direction or shape.Example 1:Finding Critical Points of a PolynomialFind all critical points ofStep 1: Identify the DomainSince (f(x)) is apolynomial, its domain isall real numbers.

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Study GuideStep 2: Find the DerivativeStep 3: Set the Derivative Equal to ZeroFactor:Solve:Step 4: Find the Corresponding Function ValuesFinal AnswerThe critical points are:Example 2:Critical Points of a Trigonometric FunctionFind all critical points of

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Study GuideStep 1: Understand the DomainThe domain is restricted to theclosed interval([0, 2π).Step 2: Find the DerivativeStep 3: Set the Derivative Equal to ZeroThis happens at:Step 4: Find the Function ValuesFinal AnswerThe critical points are:

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Study GuideKey TakeawaysTo find critical points:1.Find the derivativeof 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 functionat those (x)-values.Critical points are powerful tools that help us understand how a function behaves—and they show upeverywhere in calculus!2.Extreme Value TheoremCritical points are especially useful when we want to find themaximumandminimumvalues of afunction on a specific interval. This is where theExtreme Value Theorem (EVT)comes in.What Does the Extreme Value Theorem Say?TheExtreme Value Theoremguarantees 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)) iscontinuouson aclosed interval([a, b]), then (f(x)) hasboth amaximum and a minimum valueon that interval.In short:•Continuous function•Closed interval (includes endpoints)→ A maximum and minimummust exist.

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Study GuideHow to Use the Extreme Value TheoremTo apply the theorem, follow these steps:1.Check continuityMake sure the function is continuous on the given closed interval.2.Find all critical pointsCompute the derivative and find where:o(f'(x) = 0), oro(f'(x)) does not exist(Only include points that lieinside the interval.)3.Evaluate the functionFind the function value at:oeach critical point in the interval, andoboth endpoints of the interval.4.Compare valuesoThelargestvalue is the maximum.oThesmallestvalue is the minimum.Example 1:Trigonometric FunctionFind the maximum and minimum values ofStep 1: Check ContinuitySine and cosine are continuous, so (f(x)) is continuous on ([0, 2π]).

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Study GuideStep 2: Identify Critical PointsFrom earlier work, the critical points are:Step 3: Check the EndpointsStep 4: Compare All Values•Maximum value: (√2) at (x =π/4)•Minimum value: (-√2) at (x =5π/4)Final Answer•Maximum:(√2)•Minimum:(-√2)Note:In this example, both the maximum and minimum occur atcritical points.Example 2:Polynomial FunctionFind the maximum and minimum values ofStep 1: Check ContinuityPolynomials are continuous everywhere, so the function is continuous on ([-2, 2]).

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Study GuideStep 2: Find the DerivativeSet the derivative equal to zero:Solve:Step 3: Keep Only Points in the Interval•(x =9/4) isnotin ([-2, 2]), so we ignore it.•The only critical point in the interval is (x = 0).Step 4: Evaluate the EndpointsStep 5: Compare All Values•Maximum value:39at (x =-2)•Minimum value:–9at (x = 2)Final Answer•Maximum:39•Minimum:–9

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Study GuideImportant Note:This example shows why theclosed interval matters. Even though there was a critical point, themaximum and minimum occurred at theendpoints, not at the critical point.Key TakeawaysWhen finding absolute maximums and minimums:•Always checkcritical points AND endpoints•Only consider valuesinside the given interval•The Extreme Value Theorem guarantees the answer exists—butyou still have to find it3.Mean Value TheoremTheMean Value Theorem (MVT)connects two important ideas in calculus:•theslope of a secant line(average rate of change), and•theslope of a tangent line(instantaneous rate of change).In simple terms, the theorem says thatsomewhere between two points, the instantaneous slopematches the average slope.Figure 1The Mean Value Theorem.

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Study GuideStatement of the Mean Value TheoremIf a function (f(x)) satisfiesbothof the following conditions:•it iscontinuouson the closed interval ([a, b]), and•it isdifferentiableon the open interval ((a, b)),then there existsat least one number(c) in ((a, b)) such thatWhat Does This Mean Geometrically?•The fractionrepresents theslope of the secant linejoining the points ((a, f(a))) and ((b, f(b))).•The value (f'(c)) represents theslope of the tangent lineat some point (x = c).The Mean Value Theorem guarantees thatat least one tangent line between (a) and (b) is parallelto the secant line.A Special Case to RememberIf (f(a) = f(b)), then the slope of the secant line is zero.In this case, the theorem guaranteesat least one critical pointwhereThis idea is closely related toRolle’s Theorem.Example1:Verifying the Mean Value TheoremVerify the conclusion of the Mean Value Theorem for

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Study GuideStep 1: Check the Conditions•(f(x)) is a polynomial →continuous everywhere•Polynomials are alsodifferentiable everywhereSo, the Mean Value Theoremapplieson ([-2, 3]).Step 2: Find the Slope of the Secant LineEvaluate the function:Now compute the slope:So, the average rate of change is–2.Step 3: Find the DerivativeStep 4: Set the Derivative Equal to the Secant Slope

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