Pre-Calc Homework Solutions 294

# Pre-Calc Homework Solutions 294 - 294 Section 7.2 1 0 6...

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6. Solve x 2 2 4 x 5 x 1 6. x 2 2 5 x 2 6 5 0 ( x 2 6)( x 1 1) 5 0 x 5 6 or x 5 2 1 y 5 6 1 6 5 12 or y 5 2 1 1 6 5 5 (6, 12) and ( 2 1, 5) 7. Solve e x 5 x 1 1. From the graphs, it appears that e x is always greater than or equal to x 1 1, so that if they are ever equal, this is when e x 2 ( x 1 1) is at a minimum. } d d x } [ e x 2 ( x 1 1)] 5 e x 2 1 is zero when e x 5 1, i.e., when x 5 0. Test: e 0 5 0 1 1 5 1. So the solution is (0, 1). 8. Inspection of the graphs shows two intersection points: (0, 0), and ( p , 0). Check: 0 2 2 p ? 0 5 sin 0 5 0 and p 2 2 p 2 5 sin p 5 0. 9. Solve } x 2 2 1 x 1 } 5 x 3 . (0, 0) is a solution. Now divide by x . } x 2 2 1 1 } 5 x 2 2 5 x 4 1 x 2 x 4 1 x 2 2 2 5 0 x 2 5 } 2 1 6 ˇ 2 1 w 1 w 8 w } 5 2 2 or 1 Throw out the negative solution. x 5 6 1 y 5 x 3 5 6 1 (0, 0), ( 2 1, 2 1) and (1, 1) 10. Use the intersect function on a graphing calculator: [ 2 2, 2] by [ 2 2, 2] ( 2 0.9286, 2 0.8008), (0, 0), and (0.9286, 0.8008) Section 7.2 Exercises 1. E p 0 (1 2 cos 2 x ) dx 5 3 } 1 2 } x 2 } 1 4 } sin 2 x 4 5 } p 2 } 2. Use symmetry: 2 E p /3 0 1 } 1 2 } sec 2 t 1 4 sin 2 t 2 dt 5 E p /3 0 (sec 2 t 1 8 sin 2 t ) dt 5 3 tan t 1 4 t 2
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