BF2S Forums Math help

Can you use a graphing calculator?  that would help immensely.

Its not possible to solve algebraically (unless you know what ln(1-x^2) is), but x=0 is one solution just from looking at it, the other is likely a decimal.

Sydney wrote:

Yeah, I'm using a graphical calculator, but I can't get an exact value using that, -0.92 seems to be really close, but no cigar.

TI-83?

(2nd) (calc) (5:intersect)

the real value is an infinite decimal, so you'll never find the exact value

edit:  x=-.9165626......   y=.15991303....  is as close as you'll be able to get

Last edited by S.Lythberg (2009-05-10 12:39:36)

Sydney wrote:

Seems like I fucked up royally, the problem states 0 ≤ x ≤ 1

Therefore, I don't need to find the intersection at all to find the area

/facepalm.

you're finding the area enclosed by the two curves? just integrate the two over the range (0->1) and subtract the bottom from the top

Gawwad wrote:

Sydney wrote:

How do I find the intersection points between these two functions

e^2x and 1 - x²

So e^2x = 1 - x²

I just can't figure out how the find the two x coordinates where they intersect.

Help will be rewarded by karmaz.

Do you know logarithms?

e^2x = 1 - x²
lne^2x = ln(1 - x²)
2x = ln(1 - x²)

mm fuck me i'm too tired... Someone tell me if that's even right so far

that's right, but its no simpler to solve in that form

omg i took the ap calculus exam last week, i was zoned out the rest of the day after that shit hahaha

mcminty wrote:

To find the intersection points, you need to solve:


e^2x = 1 - x²

e^2x + x² - 1 = 0


Pretty sure you can't solve it exactly (well not easily anyway :S). You could use Newton's method a few times to approximate the root.

lol, what a joke that was, never used it again, actually, nothing past calc 1 level integrals/derivatives, even the integrals in the fourier series aren't very complicated.