Math wizards look here.

[Fluff]

Nah I solved them

[Lt Ragnarok]

O HAI THAR!

If you didn't read my facebook, I passed my first Calculus test (Limits) with 100%, so I'm off to an excellent start. ^_^  Now, I've started Derivatives, but I've also noticed that my ability to factor is hindering my ability to solve problems effectively. v_v

I'll admit it; I'm bad at factoring.  Quite frankly, I usually shortcut factoring as much as possible.  But, there are times when this becomes an absolute pain in the ass, and doing it the long way would serve me better.  However, it's been too long since I've done that, and therefore can no longer do it.  I do remember, however, that an equation with ALL ODD NUMBERS cannot be factored.

Which brings me to this:

5x^2 - 9x + 6   <----------------  Can this be factored?

The closest I got was (5x-6)(x-1), but that gives me 5x^2 - 11x + 6.  What can I do to find out if I can factor an equation?

[Ellipsis]

No I do not think that that does factorize nicely.
I'd guess that you could always use the quadratic formula though.
x=(-b(+or-)root(b^2-4ac))/(2a)

[Fluff]

First calculus test of the semester, and I totally bombed it. This must be the hardest I've ever mega-derped.

Spoiler (click to show/hide)

How to do #1, #2, and #5? (I solved #3 and #4)

[Kudo]

Well here's the easy one to fix, Problem 1.

You chained wrong when you put (2b+2g), derivative of (b^2+2gs) should be (2gv) because s is the only thing in terms of t, and s'(t) = v(t)

because v(t) = sq(b^2+2gs)

That leaves you with: .5*(2g)*sq(b^2+2gs)/sq(b^2+2gs) = g*1 = a(t), since g= a constant, a(t) = a constant.

Skipping actual numbers on Problem 2 since I'm low on time. But I'd suggest the product rule there (i). Check crit & endpoints (why they don't use "or equal to" is beyond me. This kind of question should have endpoint checking.) (ii), then find where a(t)=0. Solve with a goddamn calculator(iii). Then graph with a goddamn calculator(b).

As for Problem 5, I'd distribute out s(t) = 7*t^(5/2)- t^(7/2). Power rule separately & regularly for (a)

(b) solve for v(t) = 0, t>0.
v(t)= 35/2*t^(3/2)-7/2*t^(5/2)
= 7/2*t^(3/2)*(5-t). So t=5.

(c) again, t=5

(d) solve for a(t)=0
a(t)= 105/4*t^(1/2)-35/4*t^(3/2)
= 35/4*t^(1/2)*(3-t). So when 0<t<3.