<ul>
<li>find all the critical numbers of g(x) = 2x + sin(2x).</li>
</ul>
<p>here is what i am getting:</p>
<p>g'(x)= 2 + sin(2x) =0
sin(2x) = -2 ?????????</p>
<p>can anyone help?</p>
<p>the answer is pi(2n+1)/4 where n is an interger.</p>
<p>thank you.</p>
<p>g' = 2 + 2cos(2x) = 0
cos(2x) = -1
CN = arccos(-1) / 2</p>
<p>matty7589 did it correctly...</p>
<p>x = arccos(-1)/2
x = (n(pi))/2</p>
<p>I don't see how the answer is x=pi(2n+1)/4.</p>
<p>well, our class hasn't learned this arc thing you mentioned. Is there an alternative method of doing it?</p>
<p>and yes, that is the answer on the book. they want it in general terms, i guess.</p>
<p>arccos(x) = the angle</p>
<p>so the arccos(-1) = pi b/c the cos(pi) = -1.</p>
<p>It's the inverse function. Read this: <a href="http://www.ping.be/%7Eping1339/gonio.htm#Inverse-Trigonometri%5B/url%5D">http://www.ping.be/~ping1339/gonio.htm#Inverse-Trigonometri</a></p>
<p>wait a second!!!!!!</p>
<p>2x + sin(2x) </p>
<p>= 2x + 2sinxcosx </p>
<p>g'(x) = 2 + 2[sinx.-sinx + cosxcosx]</p>
<p>I am guessing something along those lines?</p>
<p>g'(x) = 2 + 2[sinx * -sinx + cosx * cosx]
g'(x) = 2 + 2(-sin^2(x) + cos^2(x))
g'(x) = 2 + 2(cos(2x))</p>
<p>there's a double angle formula:
cos^2(x) - sin^2(x) = cos(2x)</p>
<p>but yeah, you get the same exact thing. i would probably go with what matty did b/c it would save you some time.</p>
<p>that was a multiple choice question and that was the answer. You say there is no way to arrive at that?</p>
<p>edit: sorry, it is supposed to be pi(2n+1)/2</p>
<p>Yeah, I don't see how that works either. Even the august power of my t-89 indicates ((2n-1)pi)/2, starting with n=1.</p>
<p>sl8r: does your ti89 give you g'(x)=2 + 2cos(2x)?</p>
<p>sl8r, that is the answer!
can you explain how to get there manually?</p>
<p>that's wierd. I just took the derivative of each part of the function.</p>
<p>d/dx [2x] = 2
d/dx [sin(2x)] = 2cos(2x)
thus, g'(x) = 2 + 2cos(2x)
set equal to zero
CN: arccos (-1) / 2
hmm.</p>
<p>Since cos^-1(-1) = pi, 3pi, 5pi... (2pi -1), just divide that by 2 to get your answer.</p>
<p>Even multiples of pi don't yield -1 (i.e. cos(2pi) = 0), so you have to take them out.</p>
<p>we have never used this arccos thing in class, so there must be another way of expressing what you are doing, i assume.</p>
<p>arcsin (x) = the inverse trigonometric sine function.
this isnt really a calculus topic -- this is a pre-calculus / algebra II concept.</p>
<p>but i think the reason why you are getting the answer you are getting and george and I are getting arccos (-1) / 2 is because we are assuming a domain of [0, 2pi] whereas you are assuming an infinite domain (in which the given function is periodic, meaning it will have critical numbers repeatedly).</p>
<p>i have yet to see a trig problem in calculus that has not limited its domain, but i guess there is always a first !</p>
<p>arccos(x) = cos^-1(x).</p>
<p>If you go back to cos(2x)=-1, we could arrive at the same answer. We know that cos(pi, 3pi, 5pi...)=-1, but we're taking the cosine of 2x. Therefore, we just say that 2x = pi, 3pi, 5pi... and get x=pi/2, 3pi/2, 5pi/2... So yes, there is another way of doing this.</p>
<p>However, arccos(x) is just another way of representing this idea. The expression arccos(-1) = x is another way of saying, "cos(x) = -1," and arccos(x)=pi is another way of saying "cos(pi)=x.</p>
<p>ok thanks, I guess I should solve with the identity.</p>