# Find the points that have the common tangent line

• Apr 27th 2008, 05:21 AM
finch41
Find the points that have the common tangent line
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
• Apr 27th 2008, 06:35 AM
Sean12345
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.
• Apr 27th 2008, 06:48 AM
finch41
im pretty sure there is a common tangent line
remember this is a 4th degree function
find a graphing calculator to graph the function and you will see that 1 line can be used for 2 tangent points hence a common tangent
• Apr 28th 2008, 09:55 PM
mr fantastic
Quote:

Originally Posted by finch41
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

The easy way:

$\displaystyle f(x) = x^4 - 2x^2 - x$

$\displaystyle \Rightarrow f'(x) = 4x^3 - 4x - 1$

By symmetry it's easily seen that f'(1) = f'(-1) ......

The common tangent line is $\displaystyle y = -x - 1$. It's tangent to the curve at (1, -2) and (-1, 0).

If I have the time I might post the hard way.
• Apr 29th 2008, 01:06 PM
finch41
i have the hard way solution done by my teacher although he said he didn't like it and if i found a different solution, to show him

so yeah if you have the time, by all means
• Apr 29th 2008, 03:34 PM
mr fantastic
Quote:

Originally Posted by finch41
i have the hard way solution done by my teacher although he said he didn't like it and if i found a different solution, to show him

so yeah if you have the time, by all means

If you have a hard way solution there's no point me spending time posting another one which would probably be equivalent. Perhaps you could take the time to post the details of your hard way solution and I'll let you know if it's the same as mine.

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 ......
• Apr 29th 2008, 03:49 PM
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

-----

• Apr 29th 2008, 04:13 PM
mr fantastic
Quote:

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

-----

Mine is the same (more or less) up to and including the line I've redded.

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 $\displaystyle a = b = \frac{\sqrt{3}}{3}$ or $\displaystyle a = b = - \frac{\sqrt{3}}{3}$. But b < a => contradiction.

Therefore the solution is x = 1 and x = -1.
• Apr 29th 2008, 06:05 PM
finch41
From 3ab = 1 you get http://www.mathhelpforum.com/math-he...cea77ee2-1.gif or http://www.mathhelpforum.com/math-he...0a02a061-1.gif. But b < a => contradiction

although, cannot a = 3^1/3 and b = 3^2/3 ?
• Jan 2nd 2016, 08:24 PM
GLaw
Re: Find the points that have the common tangent line
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 $\displaystyle f(x)=x^4-2x^2-x$. If the equation of the common tangent is $\displaystyle y=mx+c$ we have

$\displaystyle m=f'(x)=4x^3-4x-1$

$\displaystyle c=f(x)-mx=x^4-2x^2-x-(4x^3-4x-1)x=-3x^4+2x^2$

We want to find two distnct points at $\displaystyle x=a$ and $\displaystyle x=b$ such that

$\displaystyle 4a^3-4a-1=4b^3-4b-1\ \Rightarrow\ a^3-b^3=a-b\ \Rightarrow\ a^2+ab+b^2=1\ \ldots\ \fbox1$

and

$\displaystyle -3a^4+2a^2=-3b^4+2b^2\ \Rightarrow\ 3(a^4-b^4)=2(a^2+b^2)$ $\displaystyle \Rightarrow\ 3(a+b)(a^2+b^2)=2(a+b)$

We will show that $\displaystyle a+b=0$. If $\displaystyle a+b\ne0$ then $\displaystyle 3(a^2+b^2)=2\ \Rightarrow\ a^2+b^2=\frac23$; substituting into $\displaystyle \fbox1$ gives $\displaystyle ab=\frac13$; also $\displaystyle (a+b)^2=a^2+b^2+2ab=\frac43$ and so $\displaystyle a+b=\frac{\pm2\sqrt3}3$. Hence $\displaystyle a,b$ are roots of the quadratic equation $\displaystyle 0=3x^2\pm2\sqrt3x+1=\left(\sqrt3x\pm1\right)^2$, i.e. $\displaystyle a=b$. We don’t want this as $\displaystyle a,b$ must be distinct.

Therefore $\displaystyle a+b=0$. Substituting $\displaystyle b=-a$ into $\displaystyle \fbox1$ gives $\displaystyle a^2=1$, i.e. $\displaystyle a=\pm1$.