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July 09, 2025, 05:18:13 am

Author Topic: jinny1's Methods Questions Thread :D  (Read 7515 times)  Share 

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jinny1

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Re: jinny1's Methods Questions Thread :D
« Reply #30 on: June 14, 2011, 08:57:57 pm »
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Sorry but very simple question here. When i diff (-x+3)^-1. I get (-x+3)^-2 .  However my calc tells me its (x-3)^-2 .   And i cannot see how that is sooo!
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Re: jinny1's Methods Questions Thread :D
« Reply #31 on: June 14, 2011, 09:02:03 pm »
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take the negative out as a factor and square it
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taiga

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Re: jinny1's Methods Questions Thread :D
« Reply #32 on: June 14, 2011, 10:12:25 pm »
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(5-3)^2 = 4
(3-5)^2 = 4

:P
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Re: jinny1's Methods Questions Thread :D
« Reply #33 on: June 15, 2011, 10:12:25 am »
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lol silly me..
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jinny1

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Re: jinny1's Methods Questions Thread :D
« Reply #34 on: July 10, 2011, 08:47:31 pm »
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hi guys i have a problem with this.



i went a little different. In the second line, i divided both sides by "x" and then squared both sides.

That would give me 4 = x+1
 thus x=3 .

What did i do wrong here??? because clearly the answer is also x=0.
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Re: jinny1's Methods Questions Thread :D
« Reply #35 on: July 10, 2011, 08:55:55 pm »
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You cant just divide both side by x to cancel it or else it will be missing a solution, take out the common factor instead
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jinny1

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Re: jinny1's Methods Questions Thread :D
« Reply #36 on: July 10, 2011, 09:05:54 pm »
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thanks.. when you say "take out the common factor instead".. is that what the worked solutions did??
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Re: jinny1's Methods Questions Thread :D
« Reply #37 on: July 10, 2011, 09:13:41 pm »
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yeah from 5th line to 6th they took out the common factor, personally from 2nd line I will move the RHS to LHS, take out x and use null factor rule
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Re: jinny1's Methods Questions Thread :D
« Reply #38 on: July 10, 2011, 09:22:09 pm »
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hi guys i have a problem with this.

(Image removed from quote.)

i went a little different. In the second line, i divided both sides by "x" and then squared both sides.

That would give me 4 = x+1
 thus x=3 .

What did i do wrong here??? because clearly the answer is also x=0.

2x(x-3)=0

would you divide by x here? :P
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jinny1

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Re: jinny1's Methods Questions Thread :D
« Reply #39 on: July 10, 2011, 09:25:13 pm »
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hi guys i have a problem with this.

(Image removed from quote.)

i went a little different. In the second line, i divided both sides by "x" and then squared both sides.

That would give me 4 = x+1
 thus x=3 .

What did i do wrong here??? because clearly the answer is also x=0.

2x(x-3)=0

would you divide by x here? :P


?? i divided by x in the second line:

2x = x(x+1)^1/2

i thought it simplied things. that becomes 2 = (x+1)^1/2

i then squared both sides.

4= x+1.

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Re: jinny1's Methods Questions Thread :D
« Reply #40 on: July 10, 2011, 10:34:22 pm »
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?? i divided by x in the second line:

2x = x(x+1)^1/2

i thought it simplied things. that becomes 2 = (x+1)^1/2

i then squared both sides.

4= x+1.
I think the reason why you can't simply divide both sides by x is because in doing that you make the assumption that x is positive.

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Re: jinny1's Methods Questions Thread :D
« Reply #41 on: July 10, 2011, 10:36:42 pm »
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?? i divided by x in the second line:

2x = x(x+1)^1/2

i thought it simplied things. that becomes 2 = (x+1)^1/2

i then squared both sides.

4= x+1.
I think the reason why you can't simply divide both sides by x is because in doing that you make the assumption that x is positive.

not quite, you're assuming that x is not 0.
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Re: jinny1's Methods Questions Thread :D
« Reply #42 on: July 10, 2011, 10:38:30 pm »
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?? i divided by x in the second line:

2x = x(x+1)^1/2

i thought it simplied things. that becomes 2 = (x+1)^1/2

i then squared both sides.

4= x+1.
I think the reason why you can't simply divide both sides by x is because in doing that you make the assumption that x is positive.

no, its because your assuming that x doesn't equal 0.

If x does equal 0(which it does) then you cannot simply divide by it
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jinny1

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Re: jinny1's Methods Questions Thread :D
« Reply #43 on: July 30, 2011, 07:29:55 pm »
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From Derrick Ha's book

Quote
Calculating an 'area between curves' is interesting in that it does not matter whether the area is
above, below or overlapping the axes. It is worthwhile justifying to yourself why this is the case.

Could someone explain why this is the case?

thanks
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Re: jinny1's Methods Questions Thread :D
« Reply #44 on: July 30, 2011, 07:52:39 pm »
+2
Lets look at a couple of cases.
1. Both curves are above the x-axis
Minusing the area under the bottom curve from the area under the top curve give just the orange area, the area between the two curves.
2. One above and one below the x-axis
The signed area above the axis is postive and the signed area below the axis is negative. We minus the area of the bottom curve, that is we minus the negative, this results in the addition of the area between the axis and the bottom curve to the top curve. That is yellow - blue (blue is negative) so the areas are added.
3. Both below the x-axis
The area between the curves and the axis are both negative, but the bottom curve's area has a greater negative. When we take this greater negative away, we add a greater value and so end up with a postive result for the area that is between the two curves
4. Partly above and partly below
Here for the sections of the top curve that are above the axis, we are just doing case 2, minusing the negative so that it adds. For the part of the top graph that is under the axis, that is the green part under the axis, we can explain it with case 3. The greater negative is minused to give the postive value, the area between the two curves.

I hope this makes sense and I haven't confused you more.
« Last Edit: July 30, 2011, 07:57:37 pm by b^3 »
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