Question

solve the following for x...

Answer 1

\[\frac{ f(x) - (-33) }{ x - 0 }\]

Answer 2

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

Answer 3

x=0

Answer 4

\[\frac{ f(x) - (-33)}{ x-0 } = f'(x)\]

Answer 5

that is the derivative formula at x=0; X=0 is your answer

Answer 6

is it? the answers says that x = 5Pi/102 and 7Pi/102

Answer 7

alright Plug in either 5pi/102 or 7pi/102 in f(x) and see if you get -33.

Answer 8

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)

Answer 9

so im not sure if the rise/run formula is the right thing to do

Answer 10

also the point (0, -33) is actually not on the curve, but this line is touching points of the curve somewhere

Answer 11

which is where answer x= 5Pi/102 and 7Pi/102 comes from

Answer 12

alright. let's see if we can write the equation of the tangent line

Answer 13

maybe?

Answer 14

with one m being rise/ivrun and the other m being the derivativer
equate the two toeeach other

Answer 15

dont think it works though

Answer 16

yea the derivative is not defined at 0

Answer 17

m = m
\[\frac{ f(x) - (-33) }{ x - 0} = f'(x)\]
\[\frac{ (36xsin(17x) + 306\sqrt{3}x^2 -33) - (-33) }{x } = 612xcos(17x) + 36\sin(17x) + 712\sqrt{3}x\]

Answer 18

i tried doing this but im still not getting the correct answer

Answer 19

did you try getting f'(x), maybe i did it wrong

Answer 20

is the x^2 in the square root?

Answer 21

no

Answer 22

oh, then I was doing it all wrong... let me start over

Answer 23

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

Answer 24

so take the limit after the rise/run formula after you factor it?

Answer 25

of the rise/run*

Answer 26

yea

Answer 27

this gives you the slope, and then what do you do to get the answer

Answer 28

\[\lim_{x \rightarrow0}36\sin(17x) + 306\sqrt{3}x^2\]

Answer 29

slope equals to 0, and then...?

Answer 30

0 = f'(x)
0 = \[0 = 612xcos(17x) + 36\sin(17x) + 712\sqrt{3}x\]

Answer 31

maybe solving for x here would be the answer?

Answer 32

the answer is 5pi/102

Answer 33

oh did you get it?

Answer 34

and 7pi/102... you took the derivative wrong

Answer 35

oh..

Answer 36

you shouldn't have 712, its 612

Answer 37

what is f'(x)?

Answer 38

ohh

Answer 39

and set them equal as you did before, 36sin(17x)+306(3)^(1/2)x=36sin(17x)+612xcos(17x)+612sqrt(3)x

Answer 40

solve for x, and thats what you get?

Answer 41

\[36\sin(17x)+306sqrt(3)x=36\sin(17x)+612xcos(17x)+612\sqrt(3)x\]

Answer 42

yea, x= (1/102)(12*k*pi+5pi) or x = (1/102)(12*k*pi-5pi)

Answer 43

n is an integer, pick n so that x is between pi/34 and 3pi/34

Answer 44

oh ok, thanks!

Answer 45

np

Answer 46

whats n?

Answer 47

hold on a second I will attach a file soon

Answer 48

there you go

Answer 49

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??

Answer 50

what is k?

Answer 51

I used k instead of n; its the same thing

Answer 52

oh ok

Answer 53

when you isolated for k , what are you finding

Answer 54

is that like the period

Answer 55

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

Answer 56

ok i finally got it

Answer 57

thanks for your help!

Answer 58

np...