# Optimization - find two points on a curve with a common tangent line?

1. Jan 9, 2010

### therest

1. The problem statement, all variables and given/known data
Find two points on curve y=x4-2x2-x that have a common tangent line.

2. Relevant equations
*the one stated above
dy/dx = 4x3-4x-1

3. The attempt at a solution
equation of a tangent line: y=mx+b

(4x3-4x-1) = m at two different points? So there are two points for which dy/dx=4x3-4x-1

I'm not sure what thinking I should be doing on this one to link the information about there being two points in the curve with the same tangent line to what I know about finding tangent lines. Will the coordinate points contain x or can I find two actual, definite points? Aren't there more than 2 places on the curve with the same tangent lines?

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2. Jan 9, 2010

### Mentallic

edit: I rechecked my work and it seems like I got the answer, but by shear luck.

Last edited: Jan 9, 2010
3. Jan 9, 2010

### Bohrok

It looks like there are only two points that have a common tangent line.

If (a, f(a)) and (b, f(b)) are the two points, you know the derivatives at the points are the same: f'(a) = f'(b)
Also, the derivative is the same as the slope of the line between the two points: m = (y2 - y1)/(x2 - x1)

You can write two equations with two unknowns to solve for the two points.

4. Jan 10, 2010

### therest

OH!

Thank you so much Bohrok! I think I know exactly what to do from there. :)

5. Jan 10, 2010

### therest

Hmm, or not. I think I'm stuck.

Here's what I did:

y=x4-2x2-x
dy/dx=4x3-4x-1=m

m=(y2-y1) / (x2-x1)=(by - ay) / (bx - ax)

--> x + y subscripts are being used to denote the x and y coordinates of points a and b which share a tangent line.

y=mx+c (c is the constant; I was already using b as a variable, sorry for confusion.)
y=(4x3-4x-1)x + c
ay=(4ax3-4ax-1)ax+c
by=(4bx3-4bx-1)bx+c

Maybe I need to resist the temptation to break it down like I would in Physics.

It seems like I'm overcomplicating the problem. Can I solve it by just finding a(x,y) and b(x,y) from those equations?

6. Jan 10, 2010

### Bohrok

Only have a few minutes right now, but I think this is what I'd do:

f'(a) = (f(b) - f(a))/(b - a)
and
f'(b) = (f(b) - f(a))/(b - a)

This is the system of equations to solve after you put in the function and its derivative. Once you know a and b, then you can start finding the actual line equation y = mx + b.
m = (by - ay) / (bx - ax), then you find the constant in the line equation.

7. Jan 11, 2010

### Mentallic

I had tried Bohrok's method earlier, but the system of equations seemed far too complicated to solve. Since I got stuck, I went back to that method again...

Once you substitute into
$$f'(a)=\frac{f(a)-f(b)}{a-b}$$
and
$$f'(b)=\frac{f(a)-f(b)}{a-b}$$

You'll end up having to solve these 2 equations:

$$4a^3-4a-1=\frac{b^4-2b^2-b-a^4+2a^2+a}{b-a}$$
and
$$4b^3-4b-1=\frac{b^4-2b^2-b-a^4+2a^2+a}{b-a}$$

Using a calculator, the solutions for (a,b) are (-1,1) and (1,1). It should be simple from here

8. Jan 12, 2010

### therest

wow, awesome! That actually makes sense! Mentallic and Bohrok, thank you so much! I ended up getting the same answers.

9. Jul 23, 2011

### aromashchenko

I have a similar problem with y=x^4-4x^3+4x^2+0.5x
(where I have to find the line, which is tangent to the curve at two points)

but I need to know how to do it without a calculator. Suggestions?

10. Jul 24, 2011

### Mentallic

As long as it doesn't seem too daunting, sure, you can do it without a calculator.

$$f'(a)=f'(b)=\frac{f(b)-f(a)}{b-a}$$

And since we're dealing with the function

$$y=x^4-4x^3+4x^2+0.5x$$

That means $$f'(a)= \frac{f(b)-f(a)}{b-a}$$ becomes $$4a^3-12a^2+8a+0.5=\frac{(b^4-4b^3+4b^2+0.5b)-(a^4-4a^3+4a^2+0.5a)}{b-a}$$

Now, on the right side, group together the power terms so in the numerator we have
$$(b^4-a^4)-4(b^3-a^3)+4(b^2-a^2)+0.5(b-a)$$ and each term has a factor of b-a in it so we can cancel that out.
Once we do that, it'll be hard to spot but you can actually divide the equation that is equal to 0 by b-a again. So now you have an equation in a and b (it's actually an ellipse) and so if we then solve for the next equation

$$f'(b)= \frac{f(b)-f(a)}{b-a}$$

Once you solve this one, you'll notice it is symmetrical to the other equation (you might even notice the symmetry before even solving it, saving you heaps of time) and so since these equations are inverses of each other, there is an obvious way of finding where they intersect each other.

11. Jul 24, 2011

### aromashchenko

thanks! that group factoring was the trick that I was missing. :-)

12. Jul 24, 2011

### Mentallic

No worries

13. Oct 20, 2013

### wishIknewmore

mentallic... how did you get you two points originally with your calculator? I'm just confused on what you tested?