What are the two points on y = x^4 - 2x^2 - x with a shared tangent line?

[k3232x]

Hi, I am having problems with the following problem. The main issue is actually starting the problem.

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

 

[cliowa]

Hi, I am having problems with the following problem. The main issue is actually starting the problem.

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

Say I'd give you a point p0=(x0,y0), where x0 and y0 are certain, fixed real values. How would you calculate the tangent line?

 

[Aspekt]

You first need to find the derivative of y = x^4 - 2x^2 - x

 

[k3232x]

I took the derivative of the equation and got:

y`= 4x^3 - 4x - 1 and put that into the equation of a tangent to the curve:

y=(4x^3 - 4x - 1)x + c ; where c is the intercept.

Then replaced y with the original equation:

x^4 - 2x^2 - x = (4x^3 -4x - 1)x + c

Then solved for the intercept:

c= -3x^4 + 2x^2

Need 2 points which have a common tangent, (a,b) and (c,d) then:

4a^3 -4a -1 = 4c^3 -4c -1 and

-3a^4 + 2a^2 = -3c^4 - 2c^2

Now here's where I am stuck. I don't know how to solve these equations which would then give me the points (a,b) and (c,d) that i need for a common tangent line.

 

[HallsofIvy]

Try writing
$$4a^3 -4a -1 = 4c^3 -4c -1$$
and
$$-3a^4 + 2a^2 = -3c^4 - 2c^2$$
as
$$4(a^3- c^3)- 4(a-c)= 0$$
and
$$-3(a^4- c^4)+ 2(a^2- c^2)= 0$$
and factor:
$$a^4- c^4= (a- c)(a^3+ a^2b+ ab^2+ b^3)$$
$$a^3- c^3= (a- c)(a^2+ ab+ b^2)$$
$$a^3- b^2= (a- c)(a+ b)$$
Obviously a= c so a- c= 0 satisfies both of your original equations. What else will?

 

[GoldPheonix]

Hi, I am having problems with the following problem. The main issue is actually starting the problem.

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

Take the derivative of the f(x).

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

That is your tangant line function, now we need to find only two points that have the same tangent line. Find the tangent line at a point. In this case, we'll find it for 1.

y' = 4(1)^3 - 4(1) - 1
y' = -1

Now, let's plug that back into the equation and try to find the other values that have the same slope.

-1 = 4x^3 - 4x - 1
0 = 4x^3 - 4x
0 = 4x(x^2 - 1)
0 = 4x(x-1)(x+1)

Which means that it will have a slope of -1 at (-1,-2); (1,-2); and (0,0).

Let's check that out.

f'(-1) = 4x^3 - 4x - 1 = -1
f'(0) = 4x^3 - 4x - 1 = -1
f'(1) = 4x^3 - 4x - 1 = -1 (From previous steps)

Confirmed.

Those points up above are points that have the same tangent line.

 

[HallsofIvy]

Take the derivative of the f(x).

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

That is your tangant line function, now we need to find only two points that have the same tangent line. Find the tangent line at a point. In this case, we'll find it for 1.

y' = 4(1)^3 - 4(1) - 1
y' = -1

Now, let's plug that back into the equation and try to find the other values that have the same slope.

-1 = 4x^3 - 4x - 1
0 = 4x^3 - 4x
0 = 4x(x^2 - 1)
0 = 4x(x-1)(x+1)

Which means that it will have a slope of -1 at (-1,-2); (1,-2); and (0,0).

Let's check that out.

f'(-1) = 4x^3 - 4x - 1 = -1
f'(0) = 4x^3 - 4x - 1 = -1
f'(1) = 4x^3 - 4x - 1 = -1 (From previous steps)

Confirmed.

Those points up above are points that have the same tangent line.

No, that gives points that have parallel tangent lines, not necessarily the same tangent line.

 

[GoldPheonix]

Oh, I see. They want the points where they have the same linear tangent line function. Well, in that case:

y + 2 = -1(x -1) => y = -x -1
y + 2 = -1(x + 1) => y = -x -3
y + 0 = -x

In which case, those points do not have equal tangent line functions. Yeah, I would use a different method to find things of equal tangent line functions.