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Pre-Calc Homework Solutions 168

# Pre-Calc Homework Solutions 168 - 168 Section 4.5 55...

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55. continued To find the critical points for 0 , x , C , we solve: 2 x 2 2 C 2 5 C ˇ C w 2 w 2 w x w 2 w 4 x 4 2 4 C 2 x 2 1 C 4 5 C 4 2 C 2 x 2 4 x 4 2 3 C 2 x 2 5 0 x 2 (4 x 2 2 3 C 2 ) 5 0 The minimum value of A ( x ) for 0 , x , C occurs at the critical point x 5 } C ˇ 2 3 w } , or x 2 5 } 3 C 4 2 } . The corresponding triangle height is a 2 mx 5 f ( x ) 2 f 9 ( x ) ? x 5 B 1 } C B } ˇ C w 2 w 2 w x w 2 w 1 } C ˇ C w Bx 2 w 2 2 w x w 2 w } 5 B 1 } C B } ! C § 2 § 2 § } 3 § C 4 § 2 } § 1 5 B 1 } C B } 1 } C 2 } 2 1 5 B 1 } B 2 } 1 } 3 2 B } 5 3 B This shows that the triangle has minimum area when its height is 3 B . Section 4.5 Linearization and Newton’s Method (pp. 220–232) Exploration 1 Approximating with Tangent Lines 1. f 9 ( x ) 5 2 x , f 9 (1) 5 2, so an equation of the tangent line is y 2 1 5 2( x 2 1) or y 5 2 x 2 1. 3. Since ( y 1 2 y 2 )(1) 5 y 1 (1) 2 y 2 (1) 5 1 2 1 5 0, this view shifts the action from the point (1, 1) to the point (1, 0). Also ( y 1 2 y 2 ) 9 (1) 5 y 1 9 (1)
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