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solve the following for x...

Mathematics
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\[\frac{ f(x) - (-33) }{ x - 0 }\]
f(x) = \[f(x) = 36xsin(17x) + 306\sqrt{3}x^2 -33\]
x=0

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Other answers:

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

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