|
http://www.zshare.net/download/202443884c626fad/
The problem is #1. I won't ask to just solve it for me, just tell me what I'm doing wrong and maybe tell me what the solution is?! xDD
What I basically did
f(x)=Cx^2
and g(x)=(1)/(x^2)
Then I set f(x) and g(x) equal to each other
so: C(x)=1/x^2
and now i multply x^2 to both sides to get Cx^2=1
Then subtract 1 to the other side and set that equal to 0.
I forgot what I did (I tried doing this a fwe days ago) but I multiply x^2/4 to (x^4-1)=0
And then I solve and I got -x^3/2 and x^3/2 as solutions..
i need to head out i'll be back later~
|
i tried to dl twice it didn't work
|
United States24770 Posts
Do you have the answer? If it's 1/4 then I can help you.
Edit: you seem to have made a mistake when setting the equations equal to each other and solving for C.
|
On October 09 2008 11:25 clazziquai wrote:http://www.zshare.net/download/202443884c626fad/The problem is #1. I won't ask to just solve it for me, just tell me what I'm doing wrong and maybe tell me what the solution is?! xDD What I basically did f(x)=Cx^2 and g(x)=(1)/(x^2) Then I set f(x) and g(x) equal to each other so: C(x)=1/x^2
Bolded part should read "so: Cx^2=1/x^2."
+ Show Spoiler +Then you can just both sides by x^2 to give you C=1/x^4 (you are solving for C, right??)
|
United States24770 Posts
On October 09 2008 12:55 vAltyR wrote:Show nested quote +On October 09 2008 11:25 clazziquai wrote:http://www.zshare.net/download/202443884c626fad/The problem is #1. I won't ask to just solve it for me, just tell me what I'm doing wrong and maybe tell me what the solution is?! xDD What I basically did f(x)=Cx^2 and g(x)=(1)/(x^2) Then I set f(x) and g(x) equal to each other so: C(x)=1/x^2 Bolded part should read "so: Cx^2=1/x^2." + Show Spoiler +Then you can just both sides by x^2 to give you C=1/x^4 (you are solving for C, right??)
Yeah this is the error I was talking about. However, he will then need to solve a quadratic in order to find the x coordinates of the intersections in terms of c (and then he can set the slope (derivative) of one graph equal to the negative reciprocal of the other slope at that x coordinate).
|
|
|
I'm on the process of working this out.
Got the derivatives of the two lines...set them equal to each other.
Solving for x. And I came across a problem x^6 = 4.
I'ts like 6:30 in the morning and I can't focus..what the hell do I do??
|
why did i ask this....wtffffffffff
LOL
wow...
|
United States24770 Posts
Ugh did you get it? Because what you said is wrong... you don't set the derivatives equal to each other since you want them orthogonal, not tangent.
|
|
|
|
|
|
EPT
