iop360
solve the following for x...
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iop360
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\[\frac{ f(x) - (-33) }{ x - 0 }\]
iop360
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f(x) = \[f(x) = 36xsin(17x) + 306\sqrt{3}x^2 -33\]
josiahh
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x=0
iop360
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\[\frac{ f(x) - (-33)}{ x-0 } = f'(x)\]
josiahh
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that is the derivative formula at x=0; X=0 is your answer
iop360
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is it? the answers says that x = 5Pi/102 and 7Pi/102
josiahh
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alright Plug in either 5pi/102 or 7pi/102 in f(x) and see if you get -33.
iop360
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the original question was find the x coordinates of all points on the curve
\[y = 36xsin(17x) + 306\sqrt{3}x^2 -33\]
with a domain of
\[\frac{ \Pi }{ 34 } < x < \frac{ 3\Pi }{ 34 }\]
where to tangent line passes through the point P(0, -33)
iop360
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so im not sure if the rise/run formula is the right thing to do
iop360
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also the point (0, -33) is actually not on the curve, but this line is touching points of the curve somewhere
iop360
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which is where answer x= 5Pi/102 and 7Pi/102 comes from
josiahh
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alright. let's see if we can write the equation of the tangent line
iop360
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maybe?
iop360
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with one m being rise/ivrun and the other m being the derivativer
equate the two toeeach other
iop360
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dont think it works though
josiahh
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yea the derivative is not defined at 0
iop360
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m = m
\[\frac{ f(x) - (-33) }{ x - 0} = f'(x)\]
\[\frac{ (36xsin(17x) + 306\sqrt{3}x^2 -33) - (-33) }{x } = 612xcos(17x) + 36\sin(17x) + 712\sqrt{3}x\]
iop360
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i tried doing this but im still not getting the correct answer
iop360
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did you try getting f'(x), maybe i did it wrong
josiahh
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is the x^2 in the square root?
iop360
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no
josiahh
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oh, then I was doing it all wrong... let me start over
josiahh
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the slope is 0; you just have to use the x in the denominator to cancel the xs in the numerator and take the limit of that as x approaches 0
iop360
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so take the limit after the rise/run formula after you factor it?
iop360
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of the rise/run*
josiahh
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yea
iop360
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this gives you the slope, and then what do you do to get the answer
iop360
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\[\lim_{x \rightarrow0}36\sin(17x) + 306\sqrt{3}x^2\]
iop360
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slope equals to 0, and then...?
iop360
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0 = f'(x)
0 = \[0 = 612xcos(17x) + 36\sin(17x) + 712\sqrt{3}x\]
iop360
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maybe solving for x here would be the answer?
josiahh
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the answer is 5pi/102
iop360
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oh did you get it?
josiahh
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and 7pi/102... you took the derivative wrong
iop360
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oh..
josiahh
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you shouldn't have 712, its 612
iop360
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what is f'(x)?
iop360
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ohh
josiahh
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and set them equal as you did before, 36sin(17x)+306(3)^(1/2)x=36sin(17x)+612xcos(17x)+612sqrt(3)x
iop360
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solve for x, and thats what you get?
josiahh
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\[36\sin(17x)+306sqrt(3)x=36\sin(17x)+612xcos(17x)+612\sqrt(3)x\]
josiahh
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yea, x= (1/102)(12*k*pi+5pi) or x = (1/102)(12*k*pi-5pi)
josiahh
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n is an integer, pick n so that x is between pi/34 and 3pi/34
iop360
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oh ok, thanks!
josiahh
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np
iop360
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whats n?
josiahh
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hold on a second I will attach a file soon
josiahh
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there you go
josiahh
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on the pdf file, it should not be 5pi + or - 12kpi; it should be 5pi+12kpi or x=-5pi+12kpi since cos(-a)=cos(a); it is 2kpi + or - a with a being 5pi/6 in our case... any questions??
iop360
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what is k?
josiahh
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I used k instead of n; its the same thing
iop360
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oh ok
iop360
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when you isolated for k , what are you finding
iop360
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is that like the period
josiahh
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no; im trying to find k. x should be in a certain range. so if x is between pi/34 and 3pi/34 then (pi/34)<x<(3pi/34). this will help you find an interval for k so that your x is in the right range or domain if you want to call it that... now k should be an integer so you look at the interval of k and decide which integers are inside that interval... if you do this for 12kpi-5pi; you should get k=1... pi/34<=(1/102)(12kpi-5pi)<=3pi/4... try it
iop360
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ok i finally got it
iop360
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thanks for your help!
josiahh
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np...