Author Topic: brightsky's Noob-tastic Question Thread (Read 16183 times) Tweet Share

brightsky · « **on:** August 25, 2010, 08:23:35 pm »

The area of the region enclosed by the graphs of $y = x^2 (x - 1)$ and $y = -x^3 + x$ . Somehow I keep getting 0 but it's obviously not...

Russ · « **Reply #1 on:** August 25, 2010, 08:29:28 pm »

I can't graph it because I can't find my TI84, but if you're using a calculator it may be summing it and getting 0 that way. Try doing the sections individually?

andy456 · « **Reply #2 on:** August 25, 2010, 08:48:49 pm »

brightsky · « **Reply #3 on:** August 25, 2010, 09:14:14 pm »

Thanks andy.

brightsky · « **Reply #4 on:** September 25, 2010, 12:10:12 pm »

Consider $f:\mathbb{R} \backslash \{0\} \to \mathbb{R}$ where $f(x) = \log_e|x|$ and $g:\left(-\infty, -\frac{5}{2}\right) \to \mathbb{R}$ where $g(x) = 2x + 5$ . Find the domain for $f(g(x))$ .

My working is that $-\infty < g(x) < 0$ , and $g(x) \neq 0$ , so $x \neq -\frac{5}{2}$ . However, $f(g(x)) = \log_e|2x + 5|$ , so $x$ remains in $\left(-\infty, -\frac{5}{2}\right)$ .

But the answer given is $x \in (-\infty, 0) \backslash \{-\frac{5}{2}\}$ .

Any help appreciated.

m@tty · « **Reply #5 on:** September 25, 2010, 12:30:09 pm »

Maximal domain of f is $\mathbb{R}\setminus\{0\}$

Range of g is $(-\infty, 0)$

Need to restrict domain of g such that it's range is a subset of the domain of f.

But it already is. So, as you said, domain stays as $\left(-\infty,-\frac{5}{2}\right)$ .

I don't know why they have given that answer, they have extended the domain past the initial constraints put on g... ??

brightsky · « **Reply #6 on:** September 25, 2010, 12:39:32 pm »

Thanks m@tty! Just for reference, the question is MAV 2006 Exam 1 Question 4 a.

stonecold · « **Reply #7 on:** September 25, 2010, 06:58:33 pm »

Yeah, I don't know what they've done there!

8039 · « **Reply #8 on:** September 25, 2010, 08:52:30 pm »

Quote from: brightsky on August 25, 2010, 08:23:35 pm

The area of the region enclosed by the graphs of $y = x^2 (x - 1)$ and $y = -x^3 + x$ . Somehow I keep getting 0 but it's obviously not...

First find the points of intersection by equaling the graphs to eachother, -0.5, 0 and 1. It would help to do a rough sketch of both graphs... then you could see that for the -0.5 to 0 $y = x^2 (x - 1)$ is on top, then from 0 to 1 is on the bottom of the other graph. I'm sure you can figure out the rest haha

brightsky · « **Reply #9 on:** October 09, 2010, 11:21:51 pm »

How do you find a write $\sin(5x)$ in terms of $\sin(x)$ and/or $\cos(x)$ ? My teacher said there is a way using De Moivre's theorem, but I don't know how...

TrueTears · « **Reply #10 on:** October 09, 2010, 11:24:42 pm »

Haha, here's some inspiration: http://vcenotes.com/forum/index.php/topic,12907.msg144763.html#msg144763

What you are finding should follow the same train of thought (almost the same!)

brightsky · « **Reply #11 on:** October 09, 2010, 11:41:35 pm »

Oh wait...me so stupid..

$\sin(5x) = \sin(x + 4x) = \sin(x)\cos(4x) + \sin(4x)\cos(x) = \sin(x) (\cos^2(2x) - \sin^2(2x)) + (2\sin(2x)\cos(2x)) \cos(x) = \sin(x) [(\cos^2(x) - \sin^2(x))]^2 - (2\sin(x)\cos(x))^2 + (4\sin(x)\cos(x))(\cos^2(x) - \sin^2(x)) \cos(x)$ then expand...

I'm still skeptical about this as I'm sure there is a more elegant solution that can be used for all simplifications of $\sin(nx)$ . My teacher said to work with $z^5$ but not sure how.

Quote from: TrueTears on October 09, 2010, 11:24:42 pm

Haha, here's some inspiration: http://vcenotes.com/forum/index.php/topic,12907.msg144763.html#msg144763

What you are finding should follow the same train of thought (almost the same!)

Thanks for that TT!

I'm not surehow effective Euler's formula would be though, or else there is some crazy algebra work?

TrueTears · « **Reply #12 on:** October 09, 2010, 11:56:54 pm »

You can keep repeatedly using it until you are happy to stop.

If you are interpreting it to be simplified to strictly sin(x) and cos(x) you should be able to still do it by repeatedly simplifying with Eulers, I haven't been bothered to try this but see where it leads you.

TrueTears · « **Reply #13 on:** October 10, 2010, 12:12:52 am »

Quote from: brightsky on October 09, 2010, 11:41:35 pm

I'm sure there is a more elegant solution that can be used for all simplifications of $\sin(nx)$ .

Yup, I'll let you do the working but i came up with this idea, obtain the cos(nx) and sin(nx) from (cos(x)+isin(x))^n using Binomial expansion, see what you get

brightsky · « **Reply #14 on:** October 10, 2010, 12:24:39 am »

Ah, smart....

Hmm..any hints on what I do with the cos(nx)?

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