Im not gonna help you with your homework mate.. sorry

Calculus...pssshhhttt. Who even knows that stuff? ^^

Steamgifts, where we will do your math homework for you. XD

LOL! I guess games and calculus don't go well together. I did that stuff too long ago to remember anything about it... but yeah, I guess the first answer is %$%@. I'm sure that was right. LOL!

Link did you ever think to google them? im a junior can't really help but google may. :)

So in life, when do you need to know the area under a curve?

What school is that?

I was actually going to try and help you... But then again, I did that stuff over 4 years ago and I don't wanna say anything wrong. Also, I'm happy I don't have to bother with that stuff anymore - although I unfortunately have to deal with physics, thanks to myself studying geography - point in case: there is no escape =P

Rewording Question #1:

limit as x -> a large number is equal to zero. Therefore I would assume that at x=9999999999999 (a large number) that the value would be close to zero (0.000001). However, there is no specific number you could pick for x that would make it zero (Infinity is not a number), so therefore the 2nd part of #1 would be false.

Hope this helps.

Bump!

Sorry, I can only do Q1 and Q3.
Q1 was solved.

For Q3, I'd find a f '(x) that fit the rule first which are: even degree (f(x) has odd degree), never give the result of 0.
It'd be something like f '(x)=x²+2. x² have even degree so x² ≥ 0. The lowest possible value of f '(x) is 2 and never be zero. (unless you use imaginary number)
Actually, it can be anything with this; a,b>0 or a,b<0
Integrate it back to find f(x), you'll get this.
Clean things up, you will get this. with a,b>0 or a,b<0 and any c.

You can also make it more complex as long as only odd degree of x exist.
Don't forget to give it a check, see if I missed anything.

Proofs are the one thing about math that I hate.

1 is fairly easy though. If F(x) = sin(x)*e^x, it crosses the x-axis, so f(x) <.0000001 at some point, and f(x) = 0 at another. Just because f(x) -> 0 as x-> infinity doesn't mean it won't cross the x-axis (potentially an infinite number of times) as x -> infinity.

2: related to the fact that, as long as f(a) doesn't equal 0, |f(a)| won't "bounce" (i.e., the limit of (d/dx)|f(x)| as x-> a is the same whether you are starting with x>a or x<. In other words, the derivative of |f(x)| is continuous for f(x) for all ranges except ones including f(x) = 0, for the same reason a piecewise function isn't differentiable.). No idea how to go about proving that, been too long since I had to bother with proofs in math.

3: f(x) = x^n + x will always result in the derivative: f'(x) = x^m + 1, where m is even. Since x raised to an even number is always positive for any real value of x, x^m + 1 is always greater than or equal to 1.

for 4, not sure how to prove it, but it revolves around the fact that there are no two numbers between 0 and -1 that, when added, = -3 (closest you're gonna get is -2, and that's with x=1 as a repeated root). That is, 1+1 does not equal 3 :D.

for 5, again don't know how to prove it, though it's not needed for this one (I imagine they don't ask for proof because the proof is way above the level of math you're doing), but even though there are an infinite number of rational numbers and an infinite number of irrational numbers, there are infinitely more irrational numbers, so (if it existed), the integral of f(x) from (0,1) = 0 (in the same way f(x) as x-> infinity = 0 in the first problem). But the integral doesn't exist.

for 6, multiplying the output of a function by a constant won't affect the input. As an example, if f(x) = int(x) from [1,5] (int(x) meaning the integer value, so int(5.34) = 5). if h(x) = kf(x), then that just means that the resulting number is different, but the domain the function is valid over is the same. The integral of h(t) in terms of f(x) = k integral(f(x). (proof: for the example, integral of f(x) from a to b = 1+2+3+4+5 = 15. integral of h(x) = k + 2k + 3k + 4k + 5k, which can be factored to k(1+2+3+4+5), or k*integral(f(x) from a to b.

Hope this helps, and sorry I couldn't help more with the proofs, but I'm tired, and I've repressed most of my knowledge of doing proofs.

hint help: use wolframalpha.com

Wow, I totally forgot calculus...

42

Probably too late, but I got nothing better to do. :(

1. Answers: YES, NO.
   According to the definition, that limit is 0 means: "For any "positive error margin" eps you choose, there is a number N such that: |f(x)-0|<eps whenever x>N. So, you can pick eps very small, say, eps=0.00000000001. There will be such a N, for sure. Just take ANY x above this N, and, voila, |f(x)|<0.00000000001, and therefore f(x)<0.0000001.
   Now, that DOES NOT NECESSARILY mean that f(x)=0 for some x. For example, you could have f(x)=1/x. Though the limit (as x->+Inf) is 0, no real value of x makes f(x)=0.

2. Since f(a)<>0, we can have f(a)>0 or f(a)<0. I will assume the first case (the other one is similar).
   Since f(x) is differentiable at a, f(x) must be continuous at a. Since f(a)>0, that means that there is some small neighborhood around a where f(x)>0 as well. In other words, there will be some "delta" such that "x in (a-delta,a+delta)" implies f(x)>0. In this neighborhood, |f(x)|=f(x) then! Since the derivative at a only depends on the behaviour of f(x) in this neighborhood, we can guarantee that the derivative of |f(x)| there is the same as f'(x); in particular, |f| is differentiable at a.
   (If f(a)<0, then |f(x)|=-f(x) in that neighborhood, and then the derivative of |f| is -f'(x) there)

3. We need something whose DERIVATIVE has no roots... Since f(x) has odd degree, I expect f'(x) to have even degree.... So, why not try something like f'(x)=A.x^(2n)+B with A and B positive? Yes, that should do! So f(x)=x^(2n+1)+x is one family of possibilities -- it has odd degree, and, since its derivative f'(x)=(2n+1)x^(2n)+1 clearly has no roots (it is always positive), we can say that f(x) has no critical points.

4. If f(x) had two roots a and b in [0,1], by Rolle (of the MVT) we would have some c between a and b such that f'(c)=(f(b)-f(a))/(b-a)=0. But, look: f'(x)=3x^2-3, whose only roots are x=1 and x=-1. None of them can be c (since 0<a<c<b<1). By contradiction, there cannot be such a pair of roots (a,b).

5. No. When taking a lower(or upper) sum, you take a partition 0=x0<x1<x2<...<xn=1. Well, ANY interval (xi,x_(i+1)) will have an irrational number in it, so the minimum of f(x) in ANY of the intervals is 0, and the corresponding Riemann sum is 0.
   Similarly, all intervals contain rational number where f=1, so the maximum of f(x) in ANY of those intervals is 1, and the Riemann Sum will be SUM(x(i+1)-x(i)).1=SUM(x(i+1)-x(i))=Total Length=1-0=1. Since all upper sums are 1 and all lower sums are 0, theintegral does not exist.

6. "Piecewise Constant" means that there is a partition a=t0<t1<t2<...<tn=b such that f(t) is constant in each interval of the form (ti,t(i+1)). But then h(t)=kf(t) will be also constant at each of those intervals, so h(t) is also piecewise constant.
   Now, write f_i for the value of f(t) when t is between ti and t(i+1). Both the lower and upper sums associated to this partition are SUM (fi . (t(i+1)-t_i)), which must then be the integral of f(t) over [a,b]. The corresponding lower and upper sums for h(t) are both SUM (kfi.(t(i+1)-t_i)), which happens to be exactly k times the corresponding sum for f. Therefore, Int [a,b] h(t) dt = k Int [a,b] f(t) dt.

7. Pick the maximum of those n numbers, say, x_M (I hope it is considered clear that such maximum exists for finite sets). Since it is the maximum, it is an upper bound, that is, x_i <=x_M for all i=1,2,3,...,n. If you take anything below x_M, you DO NOT have an upper bound for S (since x_M will be above this number, and x_M is in S). So x_M is the lowest upper bound, that is, the supremum. A similar argument shows that the minimum of those n numbers is the infimum.

I would be no help here. I had to retake pre-cal twice and when I finally passed I got a C and that was barely. Still though it was the happiest damn C of my college career. This was with hours spent in the tutor labs too lol. Calculus just did not click with me.