- #1

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- 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!

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!