A Calculus question...?

Question

Find the two points on the curve y = x^4 - 2x^3 - x that have a common tangent line.



any help is much appreciated. thanks!

Answer 1

Previous poster had the right idea but he missed a term in the derivative. Instead of,
y' = 4x^3 - 6x^2
it should be,
y' = 4x^3 - 6x^2 - 1
so his equation,
y2 - y1 = (4(x1)^3 - 6(x1)^2) * (x2 - x1)
should be,
y2 - y1 = (4(x1)^3 - 6(x1)^2 - 1) * (x2 - x1)

Also I have a sneaking suspicion there's an easier way to solve this thing.

Answer 2

Let's designate coordinates of those two points as (x1, y1) and (x2, y2). Since they are lying on the same tangent, the slope of the function must be equal at these two points. Since the function is:

y = x^4 - 2x^3 - x

the equation of the slope will be:

y' = 4x^3 - 6x^2

So for the two points in question, we can write:

4(x1)^3 - 6(x1)^2 = 4(x2)^3 - 6(x2)^2

Since these points also lie on the same straight line, whose slope is stated above, we can write:

y2 - y1 = (4(x1)^3 - 6(x1)^2) * (x2 - x1)

Finally, we can say that

y1 = (x1)^4 - 2(x1)^3 - x1
y2 = (x2)^4 - 2(x2)^3 - x2

So we have four unknowns and four equations, which should be solvable...