- #1
Samuelb88
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Homework Statement
Find the two points on the curve [tex]y=x^4-2x^2-x[/tex] that have a common tangent line.
Homework Equations
See below.
The Attempt at a Solution
[tex]y=f(x)=x^4-2x^2-x[/tex]
[tex]\frac{d}{dx}\right((x^4-2x^2-x)=4x^3-4x-1[/tex]
To my understanding, two points [tex]A,B[/tex] who have a common Tangent such that [tex]T_A=T_B[/tex]
-[tex]A(a,f(a))[/tex]
-[tex]B(b,f(b))[/tex]
[tex]T_A: y=(4a^3-4a-1)x-3a^4+2a^2[/tex]
[tex]T_B: y=(4b^3-4b-1)x-3b^4+2b^2[/tex]
By setting [tex]T_A=T_B[/tex]:
[tex]x(4a^3-4b^3-4a+4b)=3(a^4-b^4)-2(a^2-b^2)[/tex]
--> [tex]x(4(a^3-b^3)-4(a-b))=3(a^2-b^2)(a^2+b^2)-2(a^2-b^2)[/tex]
--> [tex]x((a-b)(4(a^2+ab+b^2)-4)=(a-b)(3(a+b)(a^2+b^2)-2(a+b))[/tex]
--> [tex]x=\frac{3(a+b)(a^2+b^2)-2(a+b)}{4(a^2+ab+b^2)-4)}\right(\[/tex]
Here is where I get lost. I've spent hours re-arranging the numerator and denominator through basic arithmetic operations and trying to get the denominator to cancel out in some way but to no avail. I've spent the last couple of days trying to solve this problem in various ways which is quite embarrassing because the problem seems so simple.
I've also trying setting [tex]T_A=T_B[/tex] and setting the equation equal to zero:
[tex]x(4a^3-4b^3-4a+4b)-3(a^4-b^4)+2(a^2-b^2)=0[/tex]
--> [tex]x(a-b)(4(a^2+ab+b^2)-4)-(a-b)(3(a+b)(a^2+b^2)+2(a+b))=0[/tex]
Here, from factoring the term "(a-b)" and finding the zeroes, i understand a=b, therefore a-b=0, but by substituting the term (a-b) with 0, i get 0. By substituting all a or b terms with the other, i get x=a thus x=b since a=b.
--> [tex]x(4(a^2+ab+b^2)-4)-(3(a+b)(a^2+b^2)+2(a+b))=0[/tex]
--> [tex]2x(2(a^2+ab+b^2)-2)-(a+b)(3(a^2+b^2)+2)=0[/tex]
?:(
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