AP Calculus

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Special Focus: The Fundamental Theorem of Calculus Questions 4 and 5—Round Two x Aha! By the antiderivative part of the Fundamental Theorem, if F( x) = ∫ f ( t) dt, 0 x then F′ ( x) = f ( x). Likewise, if G( x) = ∫ f ( t) dt, then G′ ( x) = f ( x). Finally, if 2 x H( x) = ∫ f ( t) dt, then H′ ( x) = f ( x). No wonder the three graphs had the same 4 behavior. Their slope functions are the same, and therefore they differ at most by a constant. The maximum value of the functions F, G, and H occurs at what x-value(s)? All three will have a local maximum value at x=2 because that is where the common slope function, f(x) changes sign from positive to negative. This is the only critical point on the interval [0,4]; the absolute maximum value must also occur at x=2. The minimum value of the functions F, G, and H occurs at what x-value(s)? For each, the only critical value occurs at x=2, where a local maximum occurs. Therefore, the minimum must occur at an endpoint. From our previous work we know that functions F and H have a value of 0 at both endpoints and that function G has a value of –8/3 at both endpoints. Therefore, the absolute minimum value for each function will occur at x=0 and at x=4. Functions F, G, and H increase on what interval(s)? Decrease on what interval(s)? The common slope function is positive on (0,2) and negative on (2,4). Thus each of the functions F, G, and H increases on [0,2] and decreases on [2,4]. Added Question Teachers can now ask a question about points of inflections that would not have been accessible to students earlier. The graphs of F, G, and H have points of inflection at what x-value(s)? We know that F′ ( x) = f ( x), G′ ( x) = f ( x), and H′ ( x) = f ( x). Therefore, F′′ ( x) = G′′ ( x) = H′′ ( x) = f ′( x) . The slope of f changes sign at x=1 and at x=3, and so the respective second derivatives change sign at these values. This is why inflection points for all three graphs occur at x=1 and at x=3. AP® Calculus: 2006–2007 Workshop Materials 55
Special Focus: The Fundamental Theorem of Calculus Question 6—Round Two From the discussion above, it is now very clear that the three functions F, G, and H differ at most by a constant and that their graphs must be (at most) vertical displacements of each other. We can even use the properties to determine those constants: and x 2 x 8 F( x) = ∫ f ( t) dt = ∫ f ( t) dt + ∫ f ( t) dt = + G( x) 0 0 2 3 x ∫ ∫ ∫ F( x) = f ( t) dt = f ( t) dt + f ( t) dt = 0 + H( x) = H( x ) . 0 0 4 This supports our earlier conjectures that F( x) = H( x) and that G( x) = F( x) − 8 3 . The above class activity also gives teachers the perfect opportunity to discuss the answer to the question many students pose when applying the antiderivative part of the Fundamental Theorem: “What happened to the constant?” This activity should help them truly understand that the constant lower limit does affect the vertical placement of the graph, but does not affect the shape of the graph. x 4 56 AP® Calculus: 2006–2007 Workshop Materials
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AP Calculus

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