Showing one positive and two neg roots

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Homework Help Overview

The discussion revolves around the polynomial equation x^3 - 12x - 7.2 = 0, focusing on demonstrating that it has one positive root and two negative roots. Participants explore various methods to analyze the roots of the equation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss using trial and error, differentiation, Descartes's rule of signs, and the intermediate value theorem to analyze the roots. Questions arise about visualizing the graph and understanding the implications of maxima and minima.

Discussion Status

There is an active exploration of different methods to confirm the nature of the roots. Some participants suggest using graphical analysis and the intermediate value theorem, while others express uncertainty about interpreting the graph correctly. No consensus has been reached yet.

Contextual Notes

Participants mention the polynomial's odd degree, which guarantees at least one real root, and discuss the continuity of the function. There are references to specific values of the function at certain points, but the discussion remains open-ended regarding the overall analysis.

synkk
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Show that the equation x^3-12x-7.2=0 has one positive and two negative roots:

I know this can be solved by trial and error, and finding f(0),f(1-4),f(-1 - -4) I have shown that It has two negative roots and one positive however I'm wondering if there is another method on how to show it has two negative and one positive root? I've found the max and min points but I am not sure what to do from there...

thanks
 
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Differentiating the function, you get

[tex]f'(x) = 3x^2 - 12[/tex]

From this you can observe that the maxima is at -2, and minima is at +2. The function is also continuous. Drawing a rough graph out, you can easily see the type of roots.
 
synkk said:
Show that the equation x^3-12x-7.2=0 has one positive and two negative roots:

I know this can be solved by trial and error, and finding f(0),f(1-4),f(-1 - -4) I have shown that It has two negative roots and one positive however I'm wondering if there is another method on how to show it has two negative and one positive root? I've found the max and min points but I am not sure what to do from there...

thanks
It's a polynomial of odd degree, so we know there is at least one real root.

Descartes's rule of signs could help. It will tell you there is one positive root and perhaps two negative roots.
 
Can you use the intermediate value theorem?

f(0) is negative. f(10) is positive so by the IVT there is a root somewhere in between...
 
Infinitum said:
Differentiating the function, you get

[tex]f'(x) = 3x^2 - 12[/tex]

From this you can observe that the maxima is at -2, and minima is at +2. The function is also continuous. Drawing a rough graph out, you can easily see the type of roots.

How can you easily see it? I've drawn a sketch and I've seen the real sketch on wolframalpha but I can't see how its two negative and one positive.
 
synkk said:
How can you easily see it? I've drawn a sketch and I've seen the real sketch on wolframalpha but I can't see how its two negative and one positive.

Well, to draw the graph you would see how the function goes at x = +/- infinity.

[tex]\lim_{x\to \infty} f(x) = \infty[/tex]

and

[tex]\lim_{x\to -\infty} f(x) = -\infty[/tex]

Now, you also know the value of f(0). And you can calculate approximate values(you only need signs) of maxima and minima. Maxima comes out to be positive and minima is negative. Now try completing the graph??
 
synkk said:
How can you easily see it? I've drawn a sketch and I've seen the real sketch on wolframalpha but I can't see how its two negative and one positive.

The points on the graph where f(x) intersects x-axis are the roots of the equation.
If you carefully look at the graph, you will see that the graph the negative x-axis at two points and the positive x-axis at one point, hence there are two negative and one positive root.
 

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