course Mth 271 9/5 1:30 If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.xxxx
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Given Solution: `a 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.** STUDENT SOLUTION AND COMMENT x/2 - x/3 > 5 (x/2)*(3/x) - (x/3) * (x/2) > 5 3x/6 - 2x/6 > 5 x/6 > 5 6 * (x/6) > 6 * 5 x>30
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Given Solution: `a 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. ** STUDENT COMMENT: I didn't have any trouble getting the factoring out I was a little confused about how to get the last line 1/2 < x < 4 I need to review this some more. after reading your solution I believe I understand how to graph and to get 1/2 < x < 4 because it would = 0 at those 2 points. INSTRUCTOR RESPONSE WITH ANOTHER ALTERNATIVE EXPLANATION: Exactly. Let me explain this in another way: The bottom line is that with an unbroken graph, you can't get from 'above the x axis' to 'below the x axis' without going through the x axis. • y values are positive above the x axis, negative below the x axis and zero on the x axis. • So you can't go from negative to positive values of y, or from positive to negative, without going through y = 0. You factored and found that y = 0 only when x = 1/2 or x = 4. So between x = 1/2 and x = 4, the graph is stuck either below or above the x axis, meaning that y is stuck being either positive or negative. x = 1 is between x = 1/2 and x = 4, and if we plug x = 1 into the function we get a negative value of y. So now we know that the function is stuck in the negative range on the interval between x = 1/2 and x = 4. So this interval is the solution to the inequality. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK