L19 - x + 2) x 2 f 00 ( x ) = 2( x + 8) x 3 Shapes:...

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Lecture 19 — Curve Sketching In the previous lecture, we saw how the derivatives of a function determine its basic shape on open in- tervals. We now want to determine how to assem- ble these shapes to form a nice sketch of the graph of an unfamiliar function. Here are some features to consider: (1) Domain (2) Intercepts (3) Asymptotes (horizontal/vertical) (4) Shapes (increasing/decreasing/concavity) (5) Local extrema, Inflection Points (6) Symmetry
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f ( x ) = 3 x 4 - 12 x 3 f 0 ( x ) = 12 x 2 ( x - 3) f 00 ( x ) = 36 x ( x - 2) Shapes: Domain/Intercepts: Asymptotes/Limits: Other Points:
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f ( x ) = 3 x 1 3 - x f 0 ( x ) = 1 - x 2 / 3 x 2 / 3 f 00 ( x ) = - 2 3 x 5 / 3 Shapes: Domain/Intercepts: Asymptotes/Limits: Other Points:
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f ( x ) = x - ln | x | + 8 x f 0 ( x ) = ( x - 4)(
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Unformatted text preview: x + 2) x 2 f 00 ( x ) = 2( x + 8) x 3 Shapes: Domain/Intercepts: Asymptotes/Limits: Other Points: f ( x ) = e-1 x f ( x ) = e-1 /x x 2 f 00 ( x ) = e-1 /x (1-2 x ) x 4 Shapes: Domain/Intercepts: Asymptotes/Limits: Other Points: f ( x ) = sin( x ) 2 + cos( x ) f ( x ) = 2 cos( x ) + 1 (2 + cos( x )) 2 f 00 ( x ) = 2 sin( x )[cos( x )-1] (2 + cos( x )) 3 Shapes: Domain/Intercepts: Asymptotes/Limits: Other Points: Try it! f ( x ) = x 2 4-x f ( x ) = x (8-x ) (4-x ) 2 f 00 ( x ) = 32 (4-x ) 3 Shapes: Domain/Intercepts: Asymptotes/Limits: Other Points:...
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L19 - x + 2) x 2 f 00 ( x ) = 2( x + 8) x 3 Shapes:...

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