Print

Please login or register.
1 Hour 1 Day 1 Week 1 Month Forever
Login with username, password and session length

[Helgoland]

Look, just divide the upper and the lower part by 2*(x-2). If that doesn't work, photograph the exercise and upload it, because that thing you posted down there is kind of weird.

[Descan]

I feel stupid because that's exactly what you're supposed to do. :/ I just thought, for some reason, that 2 was the only commonality between the two parts. :v

[Helgoland]

I know that feeling - try working on an exercise for two nights, and then have someone point out that you stated the problem the wrong way and the actual solution is two lines long. Damn I love university

[Descan]

I have another one, where it's two fractions, one minus the other.

(5x / (x-7)(x+5)) - ((x+3) / (x-7)(x+7)) is the equation

For some reason, the next step is to take the (x+5) from (x-7)(x+5), and stick it behind the terms of the second fraction to make (x+3)(x+5) / (x-7)(x+7)(x+5)

On top of that, taking the (x+7) from (x-7)(x+7) and doing the same, sticking it behind the terms of the first fraction to make 5x(x+7) / (x-7)(x+5)(x+7)

So the end result is (5x(x+7) / (x-7)(x+5)(x+7)) - ((x+3)(x+5) / (x-7)(x+7)(x+5))

And I have no idea why you do that, or why you use the second half of the denominator rather than the first. :I

Edit: I think I figured it out. Need to make the denominators match, so you modify both of them to add the characteristics of the other until they are the same. For the numerators, you modify them to reflect the changes made to the denominator.

[Another]

Yo see what part in both original denominators is common?
(x-7)
Normally when 2 ratios are summed or subtracted - you multiply each one's upper and lower parts to the full denominator of the other to get them to common denominator. Here we can immediately simplify that algorithm by not touching the common factor of denominators. (Because it would cancel later.)

[MagmaMcFry]

You do that to make both denominators equal. Basically, you take any common denominator (which in this case is (x-7)(x+7)(x+5)), and expand both fractions until their denominator matches this common denominator. Then you can add the numerators together to merge the two fractions. It's basically the same as adding rational numbers: If you want to add 4/15 and 7/10, you first need to expand them so their denominators are equal. In this case, 30 is a common multiple of both denominators, so we'll try to make them both 30. To do this, we expand 4/15 with 2, resulting in 8/30. Then we expand 7/10 with 3, resulting in 21/30. Since 4/15 = 8/30 and 7/10 = 21/30, we now know that 4/15 + 7/10 = 8/30 + 21/30 = (8+21)/30 = 29/30.

[Descan]

'Course you guys actually answer after I figured it out from some book-scrounging :v

[Vector]

. . . This is why you book-scrounge first, dear.

[Descan]

The books not well put together for learning something from scratch, it's a review for stuff I haven't done in 2-3 years. So I book-scrounged before and couldn't gain traction >___<

[Skyrunner]

Today I accidentally discovered how to do vector dotting. xD

[Pnx]

One thing I'm struggling with right now but can't really look up because of futzy internet issues (this forum still works fine, but loading most other pages takes five or six tries, which is frustrating as hell).

I've started working on stuff with the chain rule, that's the one where (f following g)¹ (x) = f¹(g(x)) * g¹(x)

I know you can chain it up with other derivative rules and such, but I'm having trouble when they're chained up in a way like, (cot(x))/(1+csc(x)), I'm not sure how to handle it.

[Jim Groovester]

Apply the quotient rule, then use the chain rule as necessary. I.E.,

f = cot x
g = 1 + csc x

f' = -csc² x
g' = -csc x cot x

(f/g)' = 1/f² (f'g - fg')

Substitute f and g in and simplify.

[Pnx]

That makes sense. Thanks.

[Descan]

I have this problem in the book. I am fairly sure it's a typo, but I want to be sure.

Transform f(x) = x^2 as a graph into y = 7f(-1/6(x-1))+1

I'm fairly certain that the y = 7f(-1/6(x-1))+1 is a typo, and needs to instead be y = 7f(-1/6(x-1))^2+1

Keep in mind, the question isn't asking for me to algebraically transform it. Otherwise I would just square-root the x^2.

To my knowledge there is no way to turn a parabola into a line through just stretching/compressing/reflecting, or moving the origin up/down/left/right.

Can someone tell me if I am wrong?

[Pnx]

Nope, pretty sure you're wrong.

The: f(x) = x^2 denotes what the function f is. The : f(-1/6(x-1)) says you're plugging in -1/6(x-1) into the function f. Then you multiply the result by 7 and add 1.

Capiche?