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Finding x-coordinates

  • Thread starter Lord Dark
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  • #1
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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 ,,
 
Last edited:

Answers and Replies

  • #2
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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
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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
33,173
4,858
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...
 

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