AP Calculus

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Special Focus: The Fundamental Theorem of Calculus Since there are no x-values where the derivative does not exist, the only critical points are x = 2, 0, –2. The domain of H is all real numbers, and H is continuous everywhere (its derivative exists everywhere). The Extreme Value Theorem does not apply because the domain is not a closed interval. We are not guaranteed the existence of any absolute extrema. Consider a sign chart: H has a local minimum at x = –2 and at x = 2 because H′ ( x) changes from negative to positive at x = –2 and at x = 2. H has a local maximum at x = 0 because H′ ( x) changes from positive to negative at x = 0. H is increasing on [ −2, 0] ∪ [ 2, ∞) . H is decreasing on ( −∞, −2] ∪ [ 0, 2] . −2 5 H( − 2) = H( 2) = 60 − 100e = −7. 032, H( 0) = 0, and H( − 4) = H( 4) > 0. −2 5 The absolute minimum value of H is H( − 2) = H( 2) = 60 − 100e = −7. 032. H has no absolute maximum value. (d) Using technology, here is a graph of y = H( x): AP® Calculus: 2006–2007 Workshop Materials 123
Special Focus: The Fundamental Theorem of Calculus Solutions to the Additional Problems 1. Take the derivative with respect to x of both sides of the given equation: d ⎛ dx ⎝ ⎜ ⎞ d ⎛ f t dt dx x 2 x x ( ) sin t 2 f ( t ) ⎠ ⎟ = − ⎝ ⎜ dt ⎞ ⎠ ⎟ . x ∫ 0 ∫ 0 Use the antiderivative part of the FTC to evaluate each side of this equation: 2 2 f ( x) = x cos( x) + 2 xsin( x) − x f ( x) . Solve this equation for f ( x): f ( x) ( 1+ x 2 )= x 2 cos( x) + 2xsin( x) , 2 x cos( x) + 2xsin( x) f ( x) = . 2 1+ x 2. (a) F'( x) = x sin x . F is increasing on [ 0, ∞ ). F is decreasing on ( −∞, 0 ]. (b) Absolute minimum value is F( 0) = 0. There is no absolute maximum value. 124 AP® Calculus: 2006–2007 Workshop Materials
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AP Calculus

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