solve the following for x...

- anonymous

solve the following for x...

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

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

- anonymous

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

- anonymous

x=0

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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)

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

maybe?

- anonymous

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

- anonymous

dont think it works though

- anonymous

yea the derivative is not defined at 0

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

is the x^2 in the square root?

- anonymous

no

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

of the rise/run*

- anonymous

yea

- anonymous

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

- anonymous

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

- anonymous

slope equals to 0, and then...?

- anonymous

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

- anonymous

maybe solving for x here would be the answer?

- anonymous

the answer is 5pi/102

- anonymous

oh did you get it?

- anonymous

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

- anonymous

oh..

- anonymous

you shouldn't have 712, its 612

- anonymous

what is f'(x)?

- anonymous

ohh

- anonymous

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

- anonymous

solve for x, and thats what you get?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

oh ok, thanks!

- anonymous

np

- anonymous

whats n?

- anonymous

hold on a second I will attach a file soon

- anonymous

there you go

##### 1 Attachment

- anonymous

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??

- anonymous

what is k?

- anonymous

I used k instead of n; its the same thing

- anonymous

oh ok

- anonymous

when you isolated for k , what are you finding

- anonymous

is that like the period

- anonymous

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)

- anonymous

ok i finally got it

- anonymous

thanks for your help!

- anonymous

np...

Looking for something else?

Not the answer you are looking for? Search for more explanations.