Solving Tangent Line Problems for Quartics & Beyond

[Jason D.]

Need Help Please! Tangent Line Problems!

Hey,

Okay, so here's the problem.
I've got this quartic:

0 = -9x^4 + 190x^3 + 840x^2 + 1992x

I need to find if this graph has two or more points with the same tangent line, and I'm at a loss. Given a point, I have no problem finding the tangent line (find slope with derivative, then just point/slope equation) but I don't know how to find two points on this graph with the same tangent line (if they exist). Please help! I'd also like to be able to do this for all graphs too, thanks!

 

[benorin]

I'll suppose you mean $$f(x) = -9x^4 + 190x^3 + 840x^2 + 1992x$$

Try the equation of a tangent line at the point (a, f(a)). Now do it again for (b, f(b)). When are they equal?

 

[Hurkyl]

Please don't multiple post.

 

[Jason D.]

sorry, I didn't know where I was supposed to put it.

Anyway,
I've got:

f(x1) - f'1(x1)x1 = f(x2) - f'(x2)x2

So I set the y intercepts equal and got that. But where do I go from there? What would I use as (a, f(a)) and (b, f(b)). I'm not sure what you mean by that.

 

[benorin]

What I meant by that is just what you did

What you have is good, but many lines may have the same y-intercept and yet not be the same. Set also the slopes equal and insist that $x_1\neq x_2$. The system thus obtained is

$$\left\{\begin{array}{cc}f(x_{1})-x_{1}f^{\prime}(x_{1})=f(x_{2})-x_{2}f^{\prime}(x_{2})\\f^{\prime}(x_{1})=f^{\prime}(x_{2}) \\ x_1\neq x_2 \end{array}\right.$$

The answer is yes! There is exactly one line tangent to f(x) at two or more points, namely

$$x_1=\frac{95}{18}-\frac{1}{18}\sqrt{42195}\mbox{ and } x_2=\frac{95}{18}+\frac{1}{18}\sqrt{42195}$$

The equation of the common tangent line is

$$y=\frac{1736927}{81}x+\frac{275062225}{2916}$$

There's a plot of f(x) and the tangent line attached at the bottom.

So I'll get you started on that system now: simplify the intercept equation while keeping in mind the derivatives (slopes) are equal to get

$$\frac{f(x_{1})-f(x_{2})}{x_{1}-x_{2}} =f^{\prime}(x_{1})=f^{\prime}(x_{2})$$

and not that $$f(x_{1})-f(x_{2})$$ has a factor of $$x_{1}-x_{2}$$ since

$$f(x_{1})-f(x_{2}) = -9x_{1}^4 + 190x_{1}^3 + 840x_{1}^2 + 1992x_{1} - \left( -9x_{2}^4 + 190x_{2}^3 + 840x_{2}^2 + 1992x_{2}\right)$$
$$= -9\left( x_{1}^4- x_{2}^4 \right) + 190\left( x_{1}^3- x_{2}^3 \right) + 840\left( x_{1}^2- x_{2}^2 \right) + 1992\left( x_{1}- x_{2} \right)$$
$$= -9\left( x_{1}^2+ x_{2}^2 \right) \left( x_{1} + x_{2} \right) \left( x_{1}- x_{2} \right) + 190\left( x_{1}- x_{2} \right) \left( x_{1}^2+x_{1}x_{2} + x_{2}^2 \right) + 840\left( x_{1} + x_{2} \right) \left( x_{1}- x_{2} \right) + 1992\left( x_{1}- x_{2} \right)$$

so that

$$\frac{f(x_{1})-f(x_{2})}{x_{1}-x_{2}} = -9\left( x_{1}^2+ x_{2}^2 \right) \left( x_{1} + x_{2} \right) + 190 \left( x_{1}^2+x_{1}x_{2} + x_{2}^2 \right) + 840\left( x_{1} + x_{2} \right) + 1992 = f^{\prime}(x_{1})=f^{\prime}(x_{2})$$

Also, $$f^{\prime}(x_{1}) = -36x_{1}^3 + 570x_{1}^2 +1680x_{1} +1992$$

and $$f^{\prime}(x_{2}) = -36x_{2}^3 + 570x_{2}^2 +1680x_{2} +1992$$

I got to do mine own homework now, sorry... see if you can get it from there. (I solved the system at the top using Maple, so I'm not yet sure that this approach will pan-out nicely, so be careful).

Oh yeah, I just put a and b instead of x1 and x2, and what I meant by that is just what you did.