I think there are no two points on the curve that share the same tangent (i.e the line is a tangent to two points on the curve), however two points may share the same gradient and hence a common tangent line and there are infinitely many of these.
There are two points on the curve y = x^4 - 2x^2 - x that have a common tangent line. Find those points.
So far I think I know what needs to be found but I cannot put the pieces together.
y-prime = 4x^3 - 4x - 1
i need a y prime that yields a slope such that the equation y=mx+b will intersect the initial curve such that the new x value will give the same derivative value as the first point (thus the same sloped tangent)
but i do not know how to do this
There's an easy way - I posted it - that boils down to intelligent guess work. Obviously the solution for x has to lie within the interval for which f'(x) = constant has more than one solution for x. A graphical approach shows how this interval would be found. x = 1 and x = -1 is an obvious candidate solution to test. insight gets rewarded ......
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
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
i hope i made no typos.... if something doesnt read clearly just ask
Continuing from that point:
(a+b)(-1-ab) + 4ab(a+b) = 0
=> (a + b)(3ab - 1) = 0
=> a + b = 0 or 3ab = 1.
From a + b = 0 you get a = 1 and b = -1 or a = -1 and b = 1.
From 3ab = 1 you get or . But b < a => contradiction.
Therefore the solution is x = 1 and x = -1.
Sorry for bumping such an old thread – I was solving a similar problem on another forum when I actually found this thread on Google.
Here’s how I do it.
Let . If the equation of the common tangent is we have
We want to find two distnct points at and such that
We will show that . If then ; substituting into gives ; also and so . Hence are roots of the quadratic equation , i.e. . We don’t want this as must be distinct.
Therefore . Substituting into gives , i.e. .