Assignment 001

course Mth 271

I am now working on the Derive program. I am not sure that I am doing what I am supposed to on it. I printed out the Initial Derive Exercise, but I can't seem to determine what I need to submit when I am finished. Is there a form or query program I need to do when I finish? As far as I can tell, for Assignments 0 and 1 this Derive Worksheet is the only thing that I have not finished.

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assignment #001

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Applied Calculus I

09-07-2006

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20:11:53

Section 0.1.28 solve x/2-x/3>5

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RESPONSE -->

first to get rid of the fractions find common denominator and the multiply the entire equation to eliminate that denominator.

the problem then would look like:

3x-2x > 30

so x >30

and the line graph would look like:

-----------(--------------------------->

30

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20:12:21

** 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 you

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RESPONSE -->

got that

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20:13:23

3x - 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 -->

ok

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20:16:32

Section 0.1 solve 2x^2+1<9x-3

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RESPONSE -->

first we would want to change the equation to say that the equation is < 0

to do that we would subtract 9x from each side and add 3 to each side to give us:

2x^2-9x+4 < 0

now we can factor it:

(2x-1)(x-4) < 0

so 2x-1 < 0 in this case x < 1/2

also x-4 < 0 so x < 4

therefore x < 1/2 or x < 4

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20:18:48

** 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 -->

it has been a while but ok I get that. I needed to state the anwer differently.

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Your work on this assignment is good. Let me know if you have questions.