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**dericteong****Member**- Registered: 2006-10-16
- Posts: 13

Can anyone help?

Stuck at here. I noticed that we need to find the value for numerator is zero and denominator is zero, then to construct a table.

How to do for following steps?

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**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

could'nt you just do:

?

ah wait, that doesnt work

*Last edited by luca-deltodesco (2006-10-23 20:23:27)*

The Beginning Of All Things To End.

The End Of All Things To Come.

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**gnitsuk****Member**- Registered: 2006-02-09
- Posts: 118

Hi,

There's a mistake in you're second line of algebra. The proof to the point you have taken it should read:

=> => => =>Now. A fraction will be negative when the numerator and denominator are of different signs. So we need to look for where

and are of different signs: =>and

=>So we see that the denominator and numerator will be of different signs when:

This is you're answer.

An easy way to see when the two functions are of different sign is to draw out a number line and above it draw a line where the numerator is less than zero, and below it draw a line where the denominator is less then zero. Then, for whichever ranges the lines are not both absent or both present, these are the areas of the number line which are your answer.

*Last edited by gnitsuk (2006-10-23 21:09:28)*

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**gnitsuk****Member**- Registered: 2006-02-09
- Posts: 118

The second post's method leads to:

Which was one of our ranges, but not the full solution.

In fact the method is incorrect as the first step involves cross multiplying by

and

But we don't know the value of x and this point and so either of these two quantities could be negative and would therefore require us to reverse the inequality sign, but we don't know if they are negative or not before we know the value of x and so we can't know if the inequality sign should be > or <

*Last edited by gnitsuk (2006-10-23 21:16:03)*

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**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

gnitsuk wrote:

But we don't know the value of x and this point and so either of these two quantities could be negative and would therefore require us to reverse the inequality sign, but we don't know if they are negative or not before we know the value of x and so we can't know if the inequality sign should be > or <

thanks for pointing out my mistake there.

The Beginning Of All Things To End.

The End Of All Things To Come.

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**polylog****Member**- Registered: 2006-09-28
- Posts: 162

gnitsuk wrote:

Hi,

I got this far but I get a different solution proceeding as follows:

We need to find the critical points of the expression, ie where it is zero or undefined:

if x = 7;it is undefined when x = 1 or x = -2.

Now looking at the signs of the expression on the intervals these conditions imply:

positive on

negative on

positive on

negative on

Therefore, the solution set is the union of the intervals

andin other words, the solution set S = {x: x < - 2} UNION {x: 1 < x < 7}

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**gnitsuk****Member**- Registered: 2006-02-09
- Posts: 118

Cool. Quiet right. I was too in a rush. For me, I prefer the method of reasoning that the quotient of the functions will be less than zero when the numerator and denominator are of differing signs. So I should have written:

Numerator is +'ve when x - 7 > 0 => x > 7

Denominator is +'ve when x < -2 or x > 1 therfore

quotient is -'ve when x < -2 or 1 < x < 7

Mitch.

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