Help with tangents?

Question

Here is the question
Find the point where the curves y=x3-3x+4 and y=3(x2-x) are tangent to each other, that is, have a common tangent line.

Answer 1

solve 3x^2-3= 6x-3
3x^2-6x =0 x=0 and x= 2
The first value is not solution
for x= 2 both y have the same value so the common point is(2,6)
and the common tangent y-6= 9(x-2)

Answer 2

Just find each derivative: y´ = 3x^2 - 3 and y´=6x - 3

Now solve the equation 3x^2 - 3=6x - 3 ==> x =0 or x =2

The point is (2,6)

For x = 0 the y is different in both functions

Answer 3

The function is y = f(x) propose we are in 2d set up. If P and Q are 2 factors on the curve The tangent at P is the proscribing posititon of secant PQ as Q has a tendency to P alongside THE CURVE. So first, the question that a secant shifted // to itself does no longer upward thrust up. 2d, secant is a line so one can no longer communicate about its "length" or a tanget at P to exist, the function must have by-product defined at P. Geometrically the curve should be gentle. X-axis is truly tangent to y = x^3 on the muse. The curves like circle and different (gentle) conic sectios provide an effect that the curve could lie totally on a similar fringe of the tangent. even though it no longer so! i wager this suffices, Madhukar Sir.