Pre-Calc Homework Solutions 127

Pre-Calc Homework Solutions 127 - Section 4.1 80 Use...

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80. Use implicit differentiation. x 2 2 y 2 5 1 } d d x } ( x 2 ) 2 } d d x } ( y 2 ) 5 } d d x } (1) 2 x 2 2 yy 9 5 0 y 9 5 } 2 2 x y } 5 } x y } y 0 5 } d d x } } x y } 5 } ( y )(1) 2 y 2 ( x )( y 9 ) } 5 5 } y 2 y 2 3 x 2 } 5 2 } y 1 3 } (since the given equation is x 2 2 y 2 5 1) At (2, ˇ 3 w ), } d dx 2 y 2 } 5 2 } y 1 3 } 5 2 } ( ˇ 1 3 w ) 3 } 5 2 } 3 ˇ 1 3 w } . Chapter 4 Applications of Derivatives Section 4.1 Extreme Values of Functions (pp. 177– 185) Exploration 1 Finding Extreme Values 1. From the graph we can see that there are three critical points: x 5 2 1, 0, 1. Critical point values: f ( 2 1) 5 0.5, f (0) 5 0, f (1) 5 0.5 Endpoint values: f ( 2 2) 5 0.4, f (2) 5 0.4 Thus f has absolute maximum value of 0.5 at x 5 2 1 and x 5 1, absolute minimum value of 0 at x 5 0, and local minimum value of 0.4 at x 5 2 2 and x 5 2. [ 2 2, 2] by [ 2 1, 1] 2. The graph of f 9 has zeros at x 5 2 1 and x 5 1 where the graph of f has local extreme values. The graph of f 9 is not defined at x 5 0, another extreme value of the graph of f .
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