Question

Find the equation of the normal line (in point slope form) to f(x)= xcos(x)+x
x=2pi

Please give a full explanation!! Thanks!

Answer 1

your idea?

Answer 2

I tried using the definition of a derivative but that got too complicated, and then Is searched a different method online and I found that the slope is 2 but 2cos2+2 seemed weird

Answer 3

f' (x ) = ?

Answer 4

f'(x)= -xsinx+cosx+1

Answer 5

so, f' ( 2pi) = ?

Answer 6

well -2pi(sin2pi)=0, and cos(2pi)+1=2, so 2?

Answer 7

yup

Answer 8

one more thing, when x = 2pi, f(2pi) = ?

Answer 9

but isn't that just the slope?

Answer 10

not yet, be patient,

Answer 11

I had trouble finding the (x,y) to put into point slope form

Answer 12

answer me, x = 2pi, f( 2pi) = ?

Answer 13

so xcosx+x --> 2pi(cos2pi) + 2pi, so 4pi?

Answer 14

so, you have a point, that is (2pi, 4pi) and you have a slope of TANGENT line is 2, And you need a NORMAL line which is perpendicular to Tangent line, right? and the slope of that NORMAL line is -1/2, got me?

Answer 15

just a question here, normal line=tangent?

Answer 16

perpendicular to tangent,

Answer 17

Yes I got you!

Answer 18

so, DAT SIT

Answer 19

so the normal line would be y-4pi=2(x-2pi)

Answer 20

I meant 1/2 for the slope lol

Answer 21

-1/2, not 1/2

Answer 22

thank you so much!!!

Answer 23

hopefully I don't make any mistake, If there is, someone points it out , please