course Mth 271 I have finished all of assignment 1, except DERIVE, which i can't find on the internet. I will try to go to VHCC next week, if you can please tell me specifically where i can find computers that have the program, as i am not very familiar w/ the campus. thanks. ©vÐÑ^ÈÎÊÑP—OÝ•×Ò¥ðI°ãÖªÔÎ¥£ÆâÈassignment #001
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13:24:25 Section 0.1.26 solve x/2-x/3>5
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RESPONSE --> first get a common denominator, which is 6. this gives you (3x/6) - (2x/6) > 5. subtract. you now have x/6 > 5. multiply both sides by 6. this gives you x>30. confidence assessment: 3
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13:25:24 It's easiest to avoid denominators where possible. So the preferred first step is to multiply both sides of the original equation by the common denominator 6: 6(x/2) - 6(x/3) = 6 * 5, which gives you3x - 2x = 6 * 5 which gives you x > 6 * 5 which simplifies to x > 30. The interval associated with this solution is 30 < x < infinity, or (30, infinity). To graph you would make an arrow starting at x = 30 and pointing to the right, indicating by an open circle that x = 30 isn't included.**
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RESPONSE --> yes, i have the graph drawn correctly in my HW from the book self critique assessment: 3
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13:30:08 Section 0.1.28 solve 2x^2+1<9x-3
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RESPONSE --> first, move terms to acheive ax^2 + bx + c form. 2x^2- 9x +4 <0 factor (2x-1) (x-4) <0 set each term equal to zero 2x-1=0, x= 1/2 x-4=0, x=4 determine test intervals: (neg. infinity, 1/2) (1/2, 4) (4, infinity). test numbers from each interval x sign <0 -1 - - no 0 - - no 1 + - yes 2 + - yes 3 + - yes 4 + + no 5 + + no therefore the solution set is 1/2
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13:32:03 The given inequality rearranges to give the quadratic 2x^2 - 9 x + 4 < 0. The left-hand side has zeros at x = .5 and x = 4, as we see by factoring [ we get (2x-1)(x-4) = 0 which is true if 2x-1 = 0 or x - 4 = 0, i.e., x = 1/2 or x = 4. ] The left-hand side is a continuous function of x (in fact a quadratic function with a parabola for a graph), and can change sign only by passing thru 0. So on each interval x < 1/2, 1/2 < x < 4, 4 < x the function must have the same sign. Testing an arbitrary point in each interval tells us that only on the middle interval is the function negative, so only on this interval is the inequality true. Note that we can also reason this out from the fact that large negative or positive x the left-hand side is greater than the right because of the higher power. Both intervals contain large positive and large negative x, so the inequality isn't true on either of these intervals. In any case the correct interval is 1/2 < x < 4. ALTERNATE BUT EQUIVALENT EXPLANATION: The way to solve this is to rearrange the equation to get 2 x^2 - 9 x + 4< 0. The expression 2 x^2 - 9 x + 4 is equal to 0 when x = 1/2 or x = 4. These zeros can be found either by factoring the expression to get ( 2x - 1) ( x - 4), which is zero when x = 1/2 or 4, or by substituting into the quadratic formula. You should be able to factor basic quadratics or use the quadratic formula when factoring fails. The function can only be zero at x = 1/2 or x = 4, so the function can only change from positive to negative or negative to positive at these x values. This fact partitions the x axis into the intervals (-infinity, 1/2), (1/2, 4) and (4, infinity). Over each of these intervals the quadratic expression can't change its sign. If x = 0 the quadratic expression 2 x^2 - 9 x + 4 is equal to 4. Therefore the expression is positive on the interval (-infinity, 1/2). The expression changes sign at x = 1/2 and is therefore negative on the interval (1/2, 4). It changes sign again at 4 so is positive on the interval (4, infinity). The solution to the equation is therefore the interval (1/2, 4), or in inequality form 1/2 < x < 4. **
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RESPONSE --> i followed the reasoning of the 2nd explanation, but i understand the 1st explanation as well. self critique assessment: 3
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