Finding x-coordinates

1. May 15, 2009

Lord Dark

1. The problem statement, all variables and given/known data
hi everyone ,, got this question and I need the idea today if possible:

Find the x-coordinate of the point on the graph of y = x^3 where
the tangent line is parallel to the secant line that cuts the curve at x = -1 and x = 1.

2. Relevant equations

3. The attempt at a solution
i got the derivative and then the tangent line equaled m=3 and then i got this equation:
y=3x+2 so my x coordinates are (1,5) (-1,-1) ,, and I think it's wrong because (-1,-1) is not parallel ,, so how to get it and then what should i do to get the coordinates ,,

Last edited: May 15, 2009
2. May 15, 2009

Staff: Mentor

Nope. Your line has slope m = 3, which is not equal to the slope of the secant line. What are the coordinates of the points on the graph of y = x^3 when x = 1 and when x = -1?

There are two points of the graph of y = x^3 with tangent lines that are parallel to the given secant line.

3. May 15, 2009

Lord Dark

yea,, got it ,, m =1 not 3 ,, and then i should y`=3x^2
3x^2=1 then I'll get the coordinates ,, x=+-sqrt(1/3) ,, thanks :D

4. May 15, 2009

Staff: Mentor

Keep in mind that 3x^2 = 1 has two solutions.

I hope that you have drawn a graph of the function you're working with...