Originally Posted by
finch41 yes i had posted the easy way i thought... i could have sworn i did make a post and now its not here... oh it must be in the other thread...
i sat down and made an accurate sketch of the graph and stumbled on the easy solution....
here is my teacher's solution: the hard way - let me know what you think
Q: y=x^4 - 2x^2 - x has two points sharing the same tangent. Find those points.
ANS: let the points be (b,f(b)) and (a,f(a))
let b < a
y prime = 4x^3 - 4x -1
so the slope of the tangent is
4a^3 - 4a -1 (equation 1) OR 4b^3 - 4b -1
4a^3 - 4a -1 = 4b^3 - 4b -1
a^3 - b^3 = a - b (after simplifying)
(a-b)(a^2 + ab + b^2) = a-b
a^2 + ab + b^2 = 1 (equation 2)
the slope of the tangent is also [f(a) - f(b)] / a-b
[a^4 - 2a^2 - a - (b^4 - 2b^2 - b) ] / a-b
= [a^4 - b^4 - 2(a^2-b^2) - (a-b)] / a-b
= [(a-b)(a^3 + a^2b + ab^2 + b^3) - 2(a-b)(a+b) - (a-b)] / (a-b)
= (a^3 + a^2b + ab^2 + b^3 - 2(a+b) - 1
= a^2(a+b) + b^2(a+b) - 2(a+b) -1
= (a+b) (a^2 + b^2 -2 ) - 1 = y prime = 4a^3 - 4a - 1 (from 1)
from equation 2 (a+b)(1-ab-2) = 4a(a^2 -1)
(a+b)(-1-ab) = 4a(-ab-b^2) from 2 again
(a+b)(-1-ab) = -4ab(a+b)
there are two cases here either a+b = 0 or a+b is not 0
if a+b is not zero then:
-1-ab = -4ab
3ab = 1 (equation N)
if N is true then substituting into equation 2 gives
a^2 + ab + b^2 = 3ab
a^2 - 2ab + b^2 = 0
(a-b)^2 = 0
a=b
but a is not equal to be as is given in the equation and was stated at the beginning
therefore a+b = 0
so we know a = -b
substitue into equation 2 gives:
(-b)^2 - b^2 + b^2 = 1
b^2 = 1
b = -1
(we began by saying b < a, and since a = -b, we need b < 0 and a > 0)
so b = -1 and a = 1
therefore the points are (-1,0) and (1,-2)
we can verify this by checking the derivative at -1 and they are equal
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i hope i made no typos.... if something doesnt read clearly just ask