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Mathematics 9 Online
OpenStudy (anonymous):

Can someone please help me, i just need this question to earn an 100 in my homework:( i don't get it at all

OpenStudy (anonymous):

OpenStudy (anonymous):

calculate where 2nd deriv is equal to 0

OpenStudy (anonymous):

i did !!!

OpenStudy (anonymous):

wait no i didn't i don't get how with trigonometry involved

OpenStudy (anonymous):

i know how to deriv but not these kind

OpenStudy (anonymous):

derivative of sin is cos derivative of cos is -sin

OpenStudy (anonymous):

so f'(x) = 5cos(x) -5sin(x) f"(x) = -5sin(x) - 5cos(x)

OpenStudy (anonymous):

so then, where, from 0 to 2pi, does f"(x) = -5sin(x) - 5cos(x) = 0?

OpenStudy (anonymous):

if x<0 or x>2pi, it doens't count

OpenStudy (anonymous):

yea i knew that, but how do i put it in terms of webassign?

OpenStudy (anonymous):

what values did you get for x?

OpenStudy (anonymous):

yea, we exclude that part!! okay:)

OpenStudy (anonymous):

no i meant i knew that sin derived is cos

OpenStudy (anonymous):

so, what value of x satisfies: -5sin(x) - 5cos(x) = 0

OpenStudy (anonymous):

none, like with my other problems i would just factor after getting the 2nd derivative and then get the values from there

OpenStudy (anonymous):

do u get what im saying?

OpenStudy (anonymous):

OpenStudy (anonymous):

thats how i did all my other examples

OpenStudy (anonymous):

trigonomic functions are a bit different from polynomials like x^2-4=0

OpenStudy (anonymous):

these values from 0, 2pie

OpenStudy (anonymous):

yea, no wonder i can't do it and keep getting it wronG:/

OpenStudy (anonymous):

well, we do some more algebra so we get sin(x)/cos(x) = 0

OpenStudy (anonymous):

sin(x)/cos(x) is tan(x)

OpenStudy (anonymous):

yes, identities are being used right?

OpenStudy (anonymous):

so find when tan(x) = 0 between 0 and 2pi

OpenStudy (anonymous):

er tan(x) =1

OpenStudy (anonymous):

not tan(x) = 0

OpenStudy (anonymous):

tan is simply sin over cosine so i just look for those values? im sorry im really bad at trig

OpenStudy (anonymous):

actually tan(x) = -1

OpenStudy (anonymous):

can u explain why please?

OpenStudy (anonymous):

you can undo the tan function with the arctan aka\[\tan^{-1} \]function

OpenStudy (anonymous):

so \[\tan^{-1}( \tan (x)) = \tan^{-1} (-1)\]which becomes\[x = \tan^{-1} (-1)\]

OpenStudy (anonymous):

you can look up values on a trig table

OpenStudy (anonymous):

okay yea i see!!! thank you for going through steps with me

OpenStudy (anonymous):

so then after that?

OpenStudy (anonymous):

plug the values you got back into the f(x), the original equation

OpenStudy (anonymous):

the -1?

OpenStudy (anonymous):

i dont get the steps at all im sorry i dont get this problem at all, i mean after you do it it makes sense but that all

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