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

dy/dx=4*x^3-4*x-1 = slope of the tangent =m.
At the two points of contact , the slopes has to be the same.

So wee try to find a slope for which 4*x^3-4*x-1 = m has real solutions.

From the theory of cubic: a*x^3 +b*x^2 +c*x +d =0
Every cubic equation with real coefficients has at least one solution x among the real numbers; this is a consequence of the intermediate value theorem. We can distinguish several possible cases using the discriminant,
Delta = 4b^3 d - b^2c^2 + 4a c^3 -18abcd +27 a^2d^2
1.If Delta <0, the equation has three distinct real roots.
2.If Delta>0, the equation has one real root and a pair of complex conjugate roots.
3.If Delta=0, then at least two roots coincide. It may be that the equation has a double root and another distinct single root, or all three roots may be the same.

Here, a=4, b=-4, c=0, d= -1-m.
We are looking for a double root of the equation in term of m. Thus, we will have two distinct points having the same m value.

Delta= 0 => 256 +256*m +432 (m+1)^2 =0
Solutions: m=-1

It is easy to see that a slope of m=-1 satisfies the problem.
So, the points are :
solve(4*x^3-4*x-1=-1,x); => x=0,1,-1
(1,-2) and (-1,0)
f(x) = x^4 -2x^2 -x
f'(x) = 4x^3 -4x - 1

f'(x) = f'(y)

4x^3 - 4x - 1 = 4y^3 - 4y -1
x(x^2 - 1) = y(y^2 -1)

so when x and y = 0 and x and y = 1
The answer is 0 and 1
This is the solution of two simultaneous equations. Unfortunately, they are non-linear equations. Your curve is:

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

You need to find a straight line that is tangent at two points. The general straight line is:

y = mx + b

Since the line must be tangent to the curve, at the tangent points (x1, y1) and (x2, y2) the slope of the line is the derivative of the curve:

m = y' = 4x^3 - 4x - 1

THe slopes must be the same for x1 and x2:

m = 4(x1)^3 - 4(x1) - 1 = 4(x2)^3 - 4(x2) -1

For any pair of x1 and x2 where this is satisfied, the slopes are equal. The twoipoints must also line along the line y = mx +b at the two point (x1, y1) and (x2, y2) so:

y1 = m x1 + b
y2 = m x2 + b


b = y1 - m x1
b = y2 - m x2

y1 - m x1 = y2 - m x2

Since the points are on teh curve:

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


y1 - m x1 = y2 - m x2
(x1)^4 -2(x1)^2 -x1 - (4(x1)^3 - 4(x1) - 1)x1 = (x2)^4 -2(x2)^2 -x2 - (4(x2)^3 - 4(x2) - 1)x2

Collecting terms:
2(x1)^2 - 3(x1)^4 = 2(x2)^2 - 3(x2)^4

Ignoring the trivial solution of x1 = x2, the even exponents lead to a solution of x1 = -x2.

Remeber that the slope must be the same for both values (+x and -x):

m(x) = 4x^3 - 4x -1
m(x) = 4x(x^2 - 1) - 1

The slope is the same for +x and -x when x^2-1 = 0

So: x1 = 1, x2 = -1
y1 = -2, y2 = 0

The answers post by the user, for information only, does not guarantee the right.

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