Find the two points on the curve that share a tangent line

[Graxum]

Homework Statement

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

Relevant Equations

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

dy/dx = 4x^3 - 4x - 1

IMPORTANT: NO CALCULATORS

I assumed two points, (a, f(a)) and (b, f(b)) where b is greater than a. Since the tangent line is shared, I did

f'(a) = f'(b):

1) 4a^3 - 4a - 1 = 4b^3 - 4b - 1
2) 4a^3 - 4a = 4b^3 - 4b
3) 4(a^3 - a) = 4(b^3 - b)
4) a^3 - a = b^3 - b
5) a^3 - b^3 = a - b
6) (a - b)(a^2 + ab + b^2) = (a - b)
7) a^2 + ab + b^2 = 1

and

(f(b) - f(a))/b - a = f'(a):

1) (b^4 - 2b^2 - b - a^4 + 2a^2 + a) / (b - a) = 4a^3 - 4a - 1
2) [(b^4 - a^4) -2(b^2 - a^2) - (b - a)] / (b - a) = 4a^3 -4a -1
3) [(b^2 + a^2)(b + a)(b - a) - 2(b + a)(b - a) - (b - a)] / (b - a) = 4a^3 - 4a - 1
4) (b^2 + a^2)(b + a) - 2(b + a) - 1 = 4a^3 - 4a - 1
5) (b + a)(b^2 + a^2 - 2) = 4a(a - 1)(a + 1)
6) From the 7th step of the 1st equation, I deduced that:
a^2 + b(a + b) = 1
6.1) b(a + b) = (1 - a)(1 + a), thus -b(a + b) = (a - 1)(a + 1)
7) Subbing in -b(a + b) for (a - 1)(a + 1) on the right side of the equation on the 5th line, I got:
(b + a)(b^2 + a^2 - 2) = -4ab(a + b)
8) b^2 + a^2 - 2 = -4ab
9) From the 7th step of the 1st equation, I deduced that:
a^2 + b^2 = 1 - ab
10) Subbing in 1 - ab for a^2 + b^2 on the left side of the equation on the 8th line, I got:
-ab - 1 = -4ab
11) 1 = 3ab, thus ab = 1/3
12) Subbing in ab for 1 - a^2 - b^2 on the right side of the equation on the 8th line, I got:
a^2 + b^2 - 2 = -4 + 4a^2 + 4b^2
13) a^2 + b^2 = 2/3

Now you would think that knowing ab = 1/3 and a^2 + b^2 = 2/3 would make things easy, however I must've made a mistake somewhere because I find multiple possible solutions, a=1 (and f(a) for the y coordinate) and 2 values for b (those being 2/3 or -1, and their respective y coordinates.) I came so far and still can't find the answer!

 

[SammyS]

Homework Statement:: Find the two points on the curve y=x^4 - 2x^2 - x that share a tangent line?
Relevant Equations:: y=x^4 - 2x^2 - x

dy/dx = 4x^3 - 4x - 1

IMPORTANT: NO CALCULATORS

I assumed two points, (a, f(a)) and (b, f(b)) where b is greater than a. Since the tangent line is shared, I did

f'(a) = f'(b):
...

and

(f(b) - f(a))/b - a = f'(a):
...

7) Subbing in -b(a + b) for (a - 1)(a + 1) on the right side of the equation on the 5th line, I got:
(b + a)(b^2 + a^2 - 2) = -4ab(a + b)
8) b^2 + a^2 - 2 = -4ab
...

Now you would think that knowing ab = 1/3 and a^2 + b^2 = 2/3 would make things easy, however I must've made a mistake somewhere because I find multiple possible solutions, a=1 (and f(a) for the y coordinate) and 2 values for b (those being 2/3 or -1, and their respective y coordinates.) I came so far and still can't find the answer!

Hello, @Graxum .



Be careful going from the result of step 7) to step 8).

You divided by $(a+b)$ . That is not legal if the value of $(a+b)$ is zero.

It's better to get zero on one side of the equation, then factor out $(a+b)$.

That should also give the possibility that $(a+b)=0$, in addition to the result you obtained.

Added in Edit:

At this point the problem is nearly solved. (Being a bit redundant, let's continue.)

$\displaystyle (b^2 + a^2 - 2 + 4ab)(a+b)=0$

So one of the two factors, $\displaystyle (b^2 + a^2 - 2 + 4ab)$ or $\displaystyle (a+b)$ must be zero.

You have already tried setting the first of these to zero. Now try the other, along with the 7th step of the 1st equation.

 

[Fred Wright]

Hint:
$$
\frac{dy}{dx}=\frac{y}{x}
$$

 

[SammyS]

Hint:
$$
\frac{dy}{dx}=\frac{y}{x}
$$

That doesn't seem to me to be correct.

 

[renormalize]

(f(b) - f(a))/b - a = f'(a):

I think keeping the problem symmetric in $a$ and $b$ will significantly simplify your algebra. Try writing your equation for the shared tangent line as:$$\frac{f(b)-f(a)}{b-a}=\frac{f^{\prime}(a)+f^{\prime}(b)}{2}$$ and repeating your derivation.

 

[pasmith]

Let the tangent line be $y = mx + c$, and define $$g(x) = f(x) - mx - c = x^4 - 2x^2 - (m+1)x - c.$$ Npw if $(a, f(a))$ is on the tangent line then $g(a) = 0$, and for the line to actually be a tangent we require $g'(a) = 0$. We obtain the same conditions for $b$. Can you see why the fact that $g$ is a polynomial of order 4 in $x$ then requires $$g(x) = (x- a)^2(x- b)^2 = x^4 - 2x^2 - (m+1)x - c?$$ Comparing coefficients of $x^3$ and $x^2$ will now give you an easier system of equations to solve for $a$ and $b$.

 

[Graxum]

Hello, @Graxum .



Be careful going from the result of step 7) to step 8).

You divided by $(a+b)$ . That is not legal if the value of $(a+b)$ is zero.

It's better to get zero on one side of the equation, then factor out $(a+b)$.

That should also give the possibility that $(a+b)=0$, in addition to the result you obtained.

Added in Edit:

At this point the problem is nearly solved. (Being a bit redundant, let's continue.)

$\displaystyle (b^2 + a^2 - 2 + 4ab)(a+b)=0$

So one of the two factors, $\displaystyle (b^2 + a^2 - 2 + 4ab)$ or $\displaystyle (a+b)$ must be zero.

You have already tried setting the first of these to zero. Now try the other, along with the 7th step of the 1st equation.

Ah i completely forgot about that!

I have 1 more question though, After the correction I got 2 answers $(a= -1/sqrt3, b= 1/sqrt3)$ from the left side and $(a=-1, b=1)$ from the right side, how do I check which of those is the correct answer? are both correct? in that case, when I plot the main equation on a graph, the 1st one seems to be completely off and those points definitely don't share a tangent line.

I wonder if I calculated something wrong somewhere?

 

[SammyS]

Ah i completely forgot about that!

I have 1 more question though, After the correction I got 2 answers $(a= -1/\sqrt 3, b= 1/\sqrt 3)$ from the left side and $(a=-1, b=1)$ from the right side, how do I check which of those is the correct answer? are both correct? in that case, when I plot the main equation on a graph, the 1st one seems to be completely off and those points definitely don't share a tangent line.

I wonder if I calculated something wrong somewhere?

Well, you could plug those values into check them.

Beyond that:
For the following equation to be true,

$\displaystyle (b^2 + a^2 - 2 + 4ab)(a+b)=0$

all that's required is that either $\displaystyle (b^2 + a^2 - 2 + 4ab)=0 \text{ or } (a+b)=0$

It looks like you found that $a+b=0$ leads to $a=-1, \ b=1$ .

That does give the correct solution.

 

[epenguin]

I thought I was fairly well up in polynomials, for an amateur, and that this would be a doddle. Instead I found it quite difficult, maybe the operative word there is 'was'. .

I didn't go through the OP's calculations, they looked too heavy and complicated. When it gets that complicated one should look for a simpler approach.

But then I realize that is even when true easier said than done sometimes. especially if you had a method which you are sure should work, so if it doesn't you might think there is some calculational error and ask for help as the OP did.

I know how to find the condition for a double root of a polynomial, I do not know a formulation for finding two double roots any more compact than finding one and then the other.

Anyway I eventually found (independently) essentially the same formulation as that of pasmith, and it all falls out in a couple of lines, with result $m = -1, c=-1$. You do not even need to consider the derivative. So I can only add, seeing is believing:

I'll worry about how general this is some other time.

 

[Steve4Physics]

For information (and given that the question is over 2 weeks old), there is a nice, short, non-calculus solution to the original problem. See the 3rd answer (near the bottom) here: https://math.stackexchange.com/questions/2458594/two-points-on-curve-that-have-common-tangent-line