Recent content by Johnnycab

omg sorry, i just wanted to find were the critical points were, but at the time i wasnt familiar with the unit circle all that much but now I am on the right track thanks to you guys, sorry if i didnt make sense i really didnt understand the 3(pi)/2 or at least didnt know what they meant but...

so if i got (-1/2) the critical points would be at 7(pi)/6 and 11(pi)/6 So for example if got (1/2) as my final answer, my critical points would be at (pi)/6, and 5(pi)/6? am i right? Also too are there 6 different circles for the 6 different trig functions? thanks sorry I...

and 2sinx+1 = (-1/2) meant to equal sinx = -(1/2) yes i did copy it down incorrectly, sorry where can i find these fields you talk of?

sorry if I am not being clear, its just that the book showed no operations for getting [3(pie)/2], [7(pie)/2], and [11(pie)/2] ^-- cause later on for other parts of the questions i need these numbers, but if i don't know where they came from then well you know

so because i got cosx=0 the critical numbers are at these points [3(pie)/2], [7(pie)/2], and [11(pie)/2] ^-- but why? is there a chart r table that i can reference from to say that when you get cosx = 0 or whatever else, your points will be at here etc

so cosx always equals those points on a closed interval, with sin? Always?

Then how did the book get the three points [3(pie)/2], [7(pie)/2], and [11(pie)/2] <--- critical points that equal zero i understand the process of which i got what i got, but then these 3 points came outta right field, any help please?

Homework Statement Find the absolute max and min of h(x) = cos(2x) - 2sinx in the closed interval [(pie)/2, 2pie] The Attempt at a Solution I got h(x) = cos(2x) - 2sinx h'(x) = -sin(2x)*2 - 2cosx <---chain rule i did :smile: 0=-2sin(2x) - 2cosx 0=[-2sin(2x)/-2] -...

yep that makes sense, thanks

sorry to cut into this post but I am working atm on a question similar to this, i haven't even seen LH rule in my book so that may be why I am going to ask this question how did you turn all those x's into 1's and 1/x in the numerator?

Yes i mean [1/(x+4)- 1/4]/x I see this equation solved out, but I am not understanding it Like after i find out that the LCD of the simple fractions in the numerator is[4(x-4)], what happened at steps 3 to 5?

thanks ok i understand multiplying the numerator and denominator by 4(x+4),and to find a LCD but the next steps are just a little confusing lim x->0 (1)[(1/x+4)-1/4]/x (2) [(1/x+4)-1/4]/x x 4(x+4) 4(x+4) = (3) [4-(x-4)]/[4x(+4)] = (4) -x/[4x(x+4)]...

Im asked to evaluate: [(1/x+4)-1/4]/x Lim -> 0 Substitution, factoring and conjugate multiplication don't work The question tells me to multipy the top and bottam by the lcd of the little fractions namely 4(x+4) --here I am a little confused, did all the book do is take both...

thanks i see what i didnt do

thank y'all :!)