Question 3 is ridiculous. The exact answer is too complicated: http://www.wolframalpha.com/input/?i=[13+-+20+x+-+x^2+-+3+x^4]%27+%3D+0

16b: no answer since the question doesn't make sense (if you do NOT ignore the typo, that is =P)

@SixWinged (sqrt(x-3)^2) != -(x-3) =P

SixWinged forgot the explanation for 16c. As t->inf exp(-0.03x) becomes exp(-inf)=0 then the formula becomes 1216/(1+75*0)=1216

SixWinged was wrong at 19. The arccos never returns negative numbers, so the answer is 3pi/5

22 and 33:

Use L'Hospital's rule which states that if lim x->0 f(x)/g(x) = 0/0 or inf/inf then it is equal to lim x->0 f'(x)/g'(x).

Then it should become obvious.

Domain for 33:

(5x^2-9x-2)=0 yields x=-1/5 or x=2

So the domain is (-inf,inf) excluding -1/5 and 2.

We all make mistakes and we are just being too nice right now to Acid :)

Also, it's l'Hopital's rule. I keep spelling it wrong because I learned it through Wolfram Alpha which has it spelled as l'Hospital's rule >_<

Heh I told you to post the problems and then I forgot to come back. I haven't read what everyone else posted so I don't know if everyone else has already solved the problems, I'm going to try anyways.

2) The simplest way would be to use ruffini, or factoring out one x and using the quadratic formula.

3) It's a polynomial, so its range is all real numbers. To get the max and min, find the first and second derivatives ( f'(x) and f''(x) ), find all x where f'(x)=0, then it's a matter of finding which are local mins/maxs and pick the appropriate one.

9) :P, (g o f)(x) = g(f(x)) = g(sqrt(x-3)) = sqrt(x-3)^2 +1 = x-3+1 = x-2

It's a linear function, it's domain and range is all real numbers, in interval notation that would be (-inf, +inf)

16)

a) Evaluate for t=0, P(0)=16

b) The easiest way to do it would be to equate the function to 750 and then find the value of t, but that is a little hardcore without a calculator, the result is 0,03t = log 56250 - log 466, or aprox. 160

c) Same as (3), have fun derivating that. However, if you are paying attention you'll notice that 1+75e^-0,03t becomes smaller and smaller towards infinity, eventually becoming irrelevant and leaving the 1216 alone, so the max is 1216.

19) Duh! 17pi/5, definition of function composition.

22) On my calculus course we assumed that lim x->0 sen(x)/x = 1. Also, it is trivial to prove that lim x->0 1/x-1 = -1. Then we simply split lim x->0 sen(x)/x^2 - x into lim x->0 sen(x)/x * lim x->0 1/x-1, which equals -1. This 'split' is justified by a limits property.

33) There's a method for this. Basically, since both polinomials are of the same degree, the limit is the division of the coefficients with the largest degree, so, the limit equals 2/5. Al.so, the domain is all real numbers except for the roots of 5x^2+9x-2

31, 34) I've never had a graphing calculator.

25) Basic trig?

29) These are two circles, and this one is easy. Replace the x^2+y^2 in the second equation with a 4 and you've reduced the problem to the intersection of the circle x^2+y^2=4 and the line x+y=2. Then you could square the second equation and reduce it to 2xy = 0, which implies that x, y or both are 0. If x is 0, the equations yield y=2, if y=0, the equations yield x=2, so, (0,2) and (2,0) are the points you're looking for.

If you plot it, you can clearly see its range is not all real numbers. =P

The function 'simplifies' to x-2, but the function still contains a square root. You cannot take the square root of negative numbers (assuming he should not take complex numbers into account).

Oh yeah :D

Hell yes you're right :P, somehow in my mind I was mixing up polynomial with linear. Auto-counterexample, x^2 is a polynomial but its range obviously is not all reals.

Yep, that's right too, f is not defined for (-inf,0) (I think we can safely assume we're not including complex numbers), so (g o f)(x) can't be defined on that interval either.

This is why you should double check your answers when you've finished your test, you never know what you might have missed, or mixed up.

Thanks again everybody, I love you guys! :)

I know a couple were stupid easy but I couldn't get my brain working again, and you guys helped me grasp logs and natural logs better. You guys are awesome and I'll make sure to ask you guys for help more often, you all explain really well, and having multiple people give different perspectives helps me wrap my head around the logic of math better. I'll be able to talk to other people once school starts, but you guys helped me do that now. I'm seriously grateful. :)