Finding X-Coordinates of Parallel Tangent and Secant Lines in y = x^3 Graph

[Lord Dark]

Homework Statement

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.

Homework Equations

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 ,,

 

[Mark44]

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.

 

[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

 

[Mark44]

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...