Thanks for that InteGrand.
I have another question. What would be the quickest method to do this?
\\\text{Find the point of intersection of the line }\\x = \begin{pmatrix}3\\2\\2\end{pmatrix} + \lambda_{1} \begin{pmatrix}-7\\-1\\-5\end{pmatrix}\\\text{And the plane }\\\omega =...
Thanks InteGrand! I had this another question:
\\\text{Suppose that }f\text{ is a continuous function on the interval [0,1] and that }f(0) = f(1)\\\text{(a) Show that }f(a) = f(a+\frac{1}{2})\text{ for some }a\in[0,\frac{1}{2}].\\\text{Hint: Let }g(x)=f(x+\frac{1}{2})-f(x)\text{ and apply the...
\\{\text{My approach:}}\\\text{We know f is continsous on }\mathbb{R}\text{ and}\lim_{x \rightarrow \infty}f(x) =\lim_{x \rightarrow -\infty}f(x) =0\\\text{Hence, regardless of smaller values of f(x), as x gets bigger or smaller, it approaches zero.}\\\text{So if }\exists\text{ }\xi\text{ such...
Thanks :) I have another question, though:
\\\text{Suppose that }f\text{ is a continuous function on }\mathbb{R}\text{ and that }\lim_{x \rightarrow \infty}f(x) =\lim_{x \rightarrow -\infty}f(x)=0 \\\text{(c) Show that if there is a real number }\xi\text{ such that }f(\xi)>0\text{ then...
Right, thanks. Do you mind (I know this sounds stupid, but) also giving a proof of the question below, because I get the derivative test, but when I differentiate it I get something very complex, so its hard to determine the points of inflexion.
Sorry, but I still don't get the question. Because when I differentiate it, I get something very complex, and it is very hard to solve it. Is there a simpler way to do this?
Is there any way this proof connects to this question I answered before it, because it says if I generalise the proof...
Thanks InteGrand, I get the proof on Wikipedia. But does that connect to my previous proof at all?
I also have another question:
\\\\\text{Find all the values of }a\text{ and }x,\text{ both in [0,2}\pi],\text{ where}\\f(x) = \cos a+2\cos(2x)+\cos(4x-a)\\\text{has a horizontal point of inflexion.}
In continuation to this question, I have another:
\\\text{Let }f\text{ be continuous on [a,b]. Prove, provided }g(x)\neq0\text{ for }x\in[a,b],\\\text{that there exists c}\in(a,b)\text{ such that:}\\\int_{a}^{b}f(t)g(t)dt=f(c)\int_{a}^{b}g(t)dt\\\text{Hint: Generalise the previous proof.}
\\\lim_{h \rightarrow 0}\frac{f(a+ph)-f(a-ph)}{h}\\\lim_{h\rightarrow 0}\frac{pf'(a+ph)+pf'(a-ph)}{1}=2pf'(a)\\\text{I don't understand where we have to make that assumption...}
Just another question:
\\\text{Let f be continuous on [a,b]. Prove there exists c}\in(a,b)\text{ such that}\\\int_{a}^{b}f(t)\text{ }dt = f(c)(b-a).\\\\\text{The way I think I can do it:}\\\text{Consider the function }F(t)\text{ where }F'(t) = f(t)\\\text{As }f(t)\text{ is continuous on [a,b]...
Thanks! Considering you gave a response at 3 in the morning!
But yeah, there are like 3 answers to the question. Here they are:
\\\text{The greatest distance is }a+2;\text{ the least distance is } \begin{cases}\sqrt{1-\frac{a^{2}}{3}} & 0\leq{a}\leq{\frac{3}{2}}\\|a-2| & a > 3/2\end{cases}.
I...
Also, how would you do this one. I could only figure out one of the solutions.
\\\text{Suppose that }a\geq0.\text{ Find the greatest and the least distances from the point }(a,0)\text{ to the ellipse}\\\frac{x^{2}}{4}+\frac{y^{2}}{1}=1
Yep, so I was just wondering how would InteGrand's explanation to part iii continue on from my working for the first two parts?
EDIT: Also, @leehuan, you're right, I mean there's no real solution to P2(x) = 0
The way I did it was I just chose an arbitrary interval [-25,0] and just went from there. Here's my working:
\\\text{(a) (i) As }p_{3}(x) = 1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}\text{ is a polynomial it is both continuous and differentiable}\\\text{everywhere. }\\\therefore\text{ We can apply...
Hey everyone, I've got another question.
\\\text{(a) (i) Show that the polynomial }p_{3}\text{ where }p_{3}(x) = 1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!},\text{ has}\text{ at least one real root}\\\text{(ii) Show that the polynomial }p_{2}\text{ where }p_{2}(x) = 1+x+\frac{x^{2}}{2},\text{ has no...
Sorry InteGrand, I was replying to leehuan's question. And after I posted it, I saw your reply. I was actually using a different (and incorrect method). I've put it below:
\frac{dt}{dV}=\frac{-1}{50\pi\sqrt{h}}\\\therefore\int_{0}^{t}dt =...
Thanks, found it. I was referring to the purple booklet. But that is literally the only proof they've given in the answers.
BTW any thoughts on the cone question?