Question

There are two points on the graph of y=x^3 where the tangent lines are parallel to y=x. Find these points

Answer 1

dy1/dx1 = dy2/dx2 ,

Answer 2

differentiate both equations implicitly with respect to x and set them equal to eachother

Answer 3

ok. I'm going to see if I can work it out

Answer 4

Will the answer be d/dx=3x^4 ?

Answer 5

3x^2, you minus for differentiation

Answer 6

so you would have dy1/dx1=3x^2 and dy2/dx2=1

Answer 7

yeah that's what I meant..lol

Answer 8

i thought as much :)

Answer 9

you can quickly see that you will indeed have two answers because of the squared x term

Answer 10

You got dy2/dx2 by the x from the parallel tangent of the line right?

Answer 11

Are the two points 3 and 6? I got 6 for the second point by finding the second derivative

Answer 12

the 2, is just a subscript, sorry for the confusion. Parrallel lines have equal slopes, and the derivative is a slope so by solving fot them being equal to eachother you will find where they are parallel.

Answer 13

\[3x^2=1 \implies x^2 = 1/3 \implies x= +- \sqrt(1/3)\]

Answer 14

Oh ok. Just looked it over myself and definitely got it. Thanks so much for your help!

Answer 15

your welcome

Answer 16

you have to plug those x values back into your original equation to get the points!

Answer 17

oh. I thought we were done..lol. So I plugged it back into the y=x^3 equation and got .19 and -.19

Answer 18

it askes for the points, so you need (x,y) we have x so we need y. we get y by plugin in the points. It should be -1/27, 1/27

Answer 19

http://www.wolframalpha.com/input/?i=%28-1%2F3%29%5E3

Answer 20

but for x didn't we get the +- square root of 1/3 and not just 1/3

Answer 21

oops your right! sorry http://www.wolframalpha.com/input/?i=%28sqrt%281%2F3%29%29%5E3

Answer 22

damnit i got an imaginary number http://www.wolframalpha.com/input/?i=%28sqrt%28-1%2F3%29%29%5E3

Answer 23

lol. Is that a good or bad thing? Clearly, I need to brush up on my math

Answer 24

I mean it's still a number