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Unit 5
10–15% of examThis unit is about using derivatives to read the shape of a graph: where a function rises and falls, where it bends, and where its highest and lowest points are. You'll also use those ideas to solve optimization problems, like finding the biggest area or the lowest cost. Expect to justify every conclusion with a theorem or a derivative test, because the AP exam grades your reasoning, not just your answer.
Longer videos that cover the whole unit. Good for a first pass or a final review.
The Mean Value Theorem says that if f is continuous on [a, b] and differentiable on (a, b), then somewhere strictly between a and b there is at least one point c where the tangent slope f′(c) equals the secant slope (f(b) − f(a))/(b − a). On the exam, you must check (and state) both conditions before you use it.
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The Extreme Value Theorem says a function that is continuous on a closed interval [a, b] must reach both an absolute maximum and an absolute minimum there. A critical point is an x-value in the domain where f′(x) = 0 or f′(x) doesn't exist, and those are the only interior places a local max or min can happen.
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A function is increasing on an interval where f′(x) > 0 and decreasing where f′(x) < 0. To find these intervals, locate the critical points, then test the sign of f′ on each piece between them.
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The First Derivative Test checks how f′ changes sign at a critical point: from positive to negative means a relative maximum, from negative to positive means a relative minimum, and no sign change means neither. A full justification names the sign change of f′, not just that f′ = 0.
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To find the absolute max and min of a continuous function on a closed interval, list the candidates (every critical point inside the interval plus both endpoints), plug each into f, and compare. The largest output is the absolute maximum and the smallest is the absolute minimum.
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A graph is concave up (curving like a cup) where f′ is increasing, which happens where f″ > 0, and concave down where f′ is decreasing, where f″ < 0. A point of inflection is where the concavity actually changes, so f″ has to change sign there; f″ = 0 alone isn't enough.
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If f′(c) = 0 and f″(c) > 0, the graph is cupped upward there, so f has a relative minimum at c; if f″(c) < 0, it has a relative maximum. If f″(c) = 0 the test tells you nothing, so go back to the First Derivative Test. A handy shortcut: if a continuous function has just one critical point on an interval and that point is a relative max (or min), it must be the absolute max (or min) on that interval too.
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You can sketch f′ from the graph of f by tracking slopes: where f is flat, f′ is zero; where f rises, f′ is positive. Working the other way, f′ and f″ tell you where f rises, falls and bends, which is enough to sketch its general shape.
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This topic ties together f, f′ and f″ in graphs, tables and equations. A common AP question gives you only the graph of f′ and asks where f has extrema or inflection points: f has a max or min where f′ changes sign, and an inflection point where f′ switches between increasing and decreasing.
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Optimization problems ask for the largest or smallest possible value of something, like area, volume, cost or distance. The first step is to write one function for the quantity you want to optimize, using a constraint (a given fact) to reduce it to a single variable, and to note the realistic domain.
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Once the function is set up, find its critical points, then use the Candidates Test or a derivative test to show which one gives the max or min. Finish by answering the question that was asked, in context and with units.
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For a curve defined by an equation in x and y (like a circle), implicit differentiation gives dy/dx in terms of both x and y. A horizontal tangent happens where the numerator of dy/dx is 0 (and the denominator isn't), and a vertical tangent where the denominator is 0 (and the numerator isn't). The second derivative d²y/dx² can then be found and used for concavity, and it may contain x, y and dy/dx.
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