equation of the normal line at a given point

I am not getting the same answer as the book for this question. it is to find the equation of the normal line at the given point

y = 2x^(3) + x^(-3) where x = -1

I solved for y getting 2 giving the point (-1,2)

taking the derivative I got y' = 6x^2 -3x^-4 and for f'(-1) I got 3

giving me the equation

y-2=3(x+1)

so for the slope of the normal line I take the negative reciprocal being -1/3

so I have

y-2 = [-1/3][x+1] I multiply everything by 3 to get [3y-6] = -1[x+1]

which gives me

x + 3y + 5 = 0

but the book has the answer down as being x+3y + 10 = 0

thank you!

Re: equation of the normal line at a given point

Looks like $\displaystyle y(-1)=-3$

Re: equation of the normal line at a given point

2x^3 + x^-3 with -1 for x

2(-1)^3 = -2 and (-1)^-3 = 1/[-1]

-2/[1/-1] is 2

or do you mean y'?

I still get 3 for y'

Re: equation of the normal line at a given point

Quote:

Originally Posted by

**Jonroberts74** I am not getting the same answer as the book for this question. it is to find the equation of the normal line at the given point

y = 2x^(3) + x^(-3) where x = -1

I solved for y getting 2 giving the point (-1,2)

taking the derivative I got y' = 6x^2 -3x^-4 and for f'(-1) I got 3

so for the slope of the normal line I take the negative reciprocal being -1/3

which gives me x + 3y + 5 = 0

but the book has the answer down as being x+3y + 10 = 0

Your point is incorrect. It should be $\displaystyle (-1,-3)$.

Re: equation of the normal line at a given point

Oh, duh haha. Silly. Thank you