Recent content by brh2113

Part (a) states Rolle's Theorem exactly. So, if you just want to know if this is true, then the answer is yes. Or do you have to prove it? What exactly are you trying to find out?

I've thought about your suggestions to go back and prove that log(a^{nx}) = n log(a^{x}) for x as any real number. If x=(1/n), then log(a^{nx}) = n log(a^{1/n}) = log (a). log(a) is the same as n(1/n)log(a), but I'm not sure if this proves that the 1/n can be brought down or if it...

I was mistakenly thinking of 1/x^{-1} = x in order to write p/q as p*(-q), but now I see that it should be p*q^{-1}, which ruins my entire plan. So backtracking, I think now I've reduce the problem to proving that f(x^{1/q}) = (1/q)f(x), because I already know that I can bring the p down...

Ah I see. Since log (a^{m+n}) = log (a^{0}) = 0 = log (a^{n}) + log (a^{m}), log (a^{n}) = log (a^{m}). This implies that nlog(a) = (-m)log(a), which means that the formula is true for all positive and negative integers, plus zero. Right? Now with r = p/q, I can write f(x^{p/q}) =...

All information, including the problem, is attached. So far I think I've proven by induction that log (a^r) = r log (a) whenever r is an integer, but I need to prove this for all rational numbers r = p/q . We're working with the functional equation that has the property that f(xy) = f(x)...

A useful trick is to write (x+2)! as (x+2)(x+1)!

Yes, r is a function of t. dr/dt is the derivative of that function with respect to t. Since dr/dt does not equal 0, r(t) is not constant. We know that dr/dt does not equal 0 because the problem says that the radius of the circle is changing with respect to time.

I think I've solved it (see attached). My only concern is that I've ignored the absolute value signs. Is this a problem? Or should I go back and work it through with two cases, one when X>0 or equal to 0 and one when X<0? That seems to me the better way, but I'm wondering if it's necessary?

I see I forgot to distribute a negative sign on the left side's derivative, but that's trivial, because as h-->0, (-h) and (h) both approach 0. Is there something else I'm missing? I've re-done the rest of the algebra, and I'm still getting 0. EDIT: I see what went wrong. I moved the h up...

I have to find the values of a and b in terms of c so that this function is differentiable. Attached is the problem and my work, but I think that there's an error somewhere in my attempt. Any advice?

Oh monotone transformation. I'm not familiar with that term, but I think I understand what you're saying to mean that each value of f^2 corresponds to one value of f? Anyways, I do agree with that inequality, and with the help of a TA in one of my classes I managed to work out the problem...

Unfortunately f is not monotonic, since it is not continuous. For example, f(a) could equal 5, and f(b) = 3, when a<b, provided that 3 and 5 are less than M. Moreover, f^2 is neither continuous nor monotonic, but I think I can assert that inequality. Still, it doesn't seem to get me anywhere...

Try treating this as a quadratic equation, except switch your B and C. Then take Ax^{2} + 2Bx + C \leq 0 Then complete the square and see what you get. (This proof is found in Apostol's Calculus)

You'll probably start with simple geometric series, which are easy to sum and determine convergence, and (if it's AP Calc) go up to Taylor Polynomials, which are used to approximate transcendental functions (among other things). Make sure that you learn all of the tests for convergence, and try...

Homework Statement Let f be a function that is integrable on [a,b] and bounded by 0 \leq f(x)\leq M for some M. Prove that f^{2} is also integrable on this interval. Homework Equations We've done many problems with step functions s(x)\leq f(x) \leq t(x), where s(x) and t(x) are...