Proof that the line intersects the curve 3 times exactly

[Dank2]

Homework Statement

p(x) = 0.2*(x-1)^5,
q(x) = 4x-6

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.

 

[member 587159]

Homework Statement

p(x) = 0.2*(x-1)^5,
q(x) = 4x-6

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.

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?

 

[Ray Vickson]

Homework Statement

p(x) = 0.2*(x-1)^5,
q(x) = 4x-6

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.

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).

 

[Dank2]

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 $h(x) = 0$). Can you continue from here?

cant get clean solutions

 

[Dank2]

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).

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.

 

[member 587159]

cant get clean solutions

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.

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

 

[Dank2]

Ah, now I see what you are doing. You should write this out explicitly in the first post. But, you are thinking in the right track then! Do you know about Rolle's theorem?

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

yes how can i use it here

 

[member 587159]

yes how can i use it here

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.

 

[Dank2]

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.

thanks

 

[Ray Vickson]

yes how can i use it here

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.

 

[Mark44]

Thread moved from Precalc section.