1. ## application of differentiation

the equation of a curve is $y$ $=$ $2$ $x^3$ $-21$ $x^2$ $+78x-98$
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....

2. Here
f '(x)=6x^2-42x+78
Now for tangent to be parrallel to yours
$6(x^2-7x+13)=6$
$x^2-7x+12=0$
Hence
$(x-3)(x-4)=0$
So the x coordinate of other point is 4