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Mathematics 14 Online
OpenStudy (anonymous):

equation of tangent line to y=xcosx at x = pi

OpenStudy (anonymous):

substitute pi into equation and got y1 = 0

OpenStudy (anonymous):

f ' (x) = cosx + x - sinx

OpenStudy (anonymous):

which slope equals -pi

OpenStudy (anonymous):

y-0 = -pi(x-pi) ?

OpenStudy (anonymous):

f'(x) = cos(x) + -xsin(x) I believe

OpenStudy (anonymous):

no its not.

OpenStudy (anonymous):

i thought so too but according to mathway.com its cosx - sinx

OpenStudy (anonymous):

Your slope is -1.

OpenStudy (anonymous):

can you show me how ?

OpenStudy (anonymous):

cos(pi) =-1 sin (pi)=0 thus -1 +0 = -1

OpenStudy (anonymous):

its -xsinx

OpenStudy (anonymous):

-x sinx same thing would be pi times sin(x) or pi times 0.

OpenStudy (anonymous):

oh right.

OpenStudy (anonymous):

okay so for the equation

OpenStudy (anonymous):

y1 = 0 right ?

OpenStudy (anonymous):

y-xcosx= 1(x-pi) Now solve for y.

OpenStudy (anonymous):

its negative 1 not positive 1

OpenStudy (anonymous):

Right

OpenStudy (anonymous):

so is it-1 - pi ?

OpenStudy (anonymous):

y-xcos(x)=-1(x-pi) Solve for y.

OpenStudy (anonymous):

y = -x ?

OpenStudy (anonymous):

OK solve the right side of the equation first. -1(x-pi) = what>

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