Thread: application of differentiation

the equation of a curve is
Find the gradient at the point (3,1). Find the x-coordinate of the point at which the tangent of the curve is parallel to the tangent at (3,1).

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i can solve the first part of the question.....answer is 6....
the second part of the question is the one i have problem with.....it sounds very confusing....

Here
f '(x)=6x^2-42x+78
Now for tangent to be parrallel to yours


Hence

So the x coordinate of other point is 4