# Proof that the line intersects the curve 3 times exactly

1. Dec 20, 2016

### Dank2

1. The problem statement, all variables and given/known data
p(x) = 0.2*(x-1)^5,
q(x) = 4x-6

3. The attempt at a solution
I took the diffrence h(x) = p(x) - q(x)
h'(x) = ((x-1)^4) - 4
got two solutions for h'(x)=0.

2. Dec 20, 2016

### Math_QED

I believe you are doing something wrong here. You have to prove that the line $y_1 = 4x - 6$ intersects with $y_2 = \frac{1}{5}(x-1)^5$. What does this mean? You have to find the points $x$ for which $y_1$ = $y_2$. Thus, we set $y_1 = y_2$ and solve for $x$.

$y_1 = y_2$
$\iff y_1 - y_2 = 0$
$\iff 4x - 6 - (\frac{1}{5}(x-1)^5)= 0$

There's no need to use derivatives! (In your notation, you have to find the points $x$ such that $h(x) = 0$, not $h'(x) = 0$). Can you continue from here?

3. Dec 20, 2016

### Ray Vickson

What does this have to do with the original question? Finding the stationary points of h(x) is not the same thing as finding the zeros of h(x).

4. Dec 20, 2016

### Dank2

cant get clean solutions

Last edited by a moderator: Dec 20, 2016
5. Dec 20, 2016

### Dank2

maybe i can conclude about the max solutions that is possible. if only at two points h'(x) = 0 , then intuitively i can see why there can't be more than 3 solutions. can't see how to proof it though.

6. Dec 20, 2016

### Math_QED

Ah, now I see what you are doing. You should write this out explicitly in the first post. Yes, it might be impossible to find the zeros of the difference between those 2 functions.

But, you are thinking in the right track when you suggest using derivatives! Do you know about Rolle's theorem?

https://en.wikipedia.org/wiki/Rolle's_theorem

7. Dec 20, 2016

### Dank2

yes how can i use it here

8. Dec 20, 2016

### Math_QED

Suppose there are more than 3 zeros and use the fact that there are only 2 zeros for the derivative to derive a contradiction with Rolle's theorem.

EDIT: I have to add that, with this approach, you only proved that it has a maximum of 3 zeros. You also need to prove that it has exactly 3 zeros.

Last edited: Dec 20, 2016
9. Dec 20, 2016

### Dank2

thanks

10. Dec 20, 2016

### Ray Vickson

If h'(x) has two zeros it is possible that h(x) also has only two (real) zeros: one isolated zero, and one zero of multiplicity two. (Of course, one can check that this example will not exhibit that behavior, because one can check that no stationary point is a (double) zero of h(x).) However, the argument certainly does eliminate the possibility of more than 3 zeros.

11. Dec 20, 2016