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