Tangent line

1. Oct 25, 2006

k3232x

Hi, im 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.

2. Oct 25, 2006

cliowa

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?

3. Oct 25, 2006

Aspekt

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

4. Oct 26, 2006

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.

5. Oct 26, 2006

HallsofIvy

Staff Emeritus
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?

Last edited: Oct 26, 2006
6. Oct 26, 2006

GoldPheonix

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.

7. Oct 27, 2006

HallsofIvy

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

8. Oct 28, 2006

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.