Differentiation

footprints

Find the equation of the tangent to the curve [tex]y=x^3-7x^2+14x-8[/tex] at the point where [tex]x = 1[/tex]. [tex]\text{Answer: }y = 3x -3[/tex]
Find the x-coordinate of the point at which the tangent is parallel to the tangent at [tex]x = 1[/tex].

I need help on the second part.

arildno

Well, let x* be the point you seek.
What do you know of x*?
Do you agree that we must have y'(x*)=3?
That is, the derivatives of y(x) must be equal at x=1 and x=x*

footprints

Yes.
I think I get it. But I still don't know how to get the answer.

arildno

Well, do you agree that what you need to solve is the equation (written with "x"):
[tex]3=3x^{2}-14x+14[/tex]
This can be rewritten as:
[tex]3(x+1)(x-1)-14(x-1)=0\to(3(x+1)-14)(x-1)=0[/tex]
What must then x* be?

footprints

Oh! Thank you!