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Find the number of distinct real roots

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anemone

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Feb 14, 2012
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Let $f(x)=x^3-3x+1$. Find the number of distinct real roots of the equation $f(f(x))=0$.
 

jacks

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Apr 5, 2012
226
Just asking

I am getting Total no. of real solution is $ = 7$

Is is Right or not

Thanks
 

mathbalarka

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Mar 22, 2013
573
Quite a nice bit of a problem you've got there. Here's my solution --

Finding the real roots of \(\displaystyle f(f(x)) = 0\) is equivalent to finding the ones of \(\displaystyle f(x) = x_i\) where \(\displaystyle x_i\) for i = 1, 2, 3 are the (real) roots of f(x).

We see that the roots of \(\displaystyle f(x)\) are all real as the discriminant is positive. An inspection of \(\displaystyle f(x) = x^3 - 3x + 1\) can be considered at the intervals \(\displaystyle [-2, -1] \cup [-1, 0] \cup [0, 1] \cup [1, 2]\) to find out that three roots lie precisely in between the first, third and the fourth interval.

Note that for the equation \(\displaystyle f(x) = x_i\) to have all real root, the discriminant must be positive, implying the \(\displaystyle -2 \leq 1 - x_i \leq 2\). The first root lies between [-2, -1], so \(\displaystyle 1 - x_1\) lies in between 2 and 3 and hence the discriminant comes down to a negative value, so one real root here exists and the other twos are complex conjugates of each other in \(\displaystyle \mathbb{C}\).

For the second root, it lies between 0 and 1, so \(\displaystyle 1 - x_2\) lies in between 2 and -2 obviously enough. The same applies for the third root. So we have 3 real roots each for these two.

So, the total number of real roots are 1 + 3 + 3 = 7.


Balarka
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anemone

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Feb 14, 2012
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Just asking

I am getting Total no. of real solution is $ = 7$

Is is Right or not

Thanks
Yes, 7 is the correct answer!:)

And thanks to you, Balarka, for participating and for your correct answer with your nice method!
 
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anemone

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Feb 14, 2012
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Finding the real roots of \(\displaystyle f(f(x)) = 0\) is equivalent to finding the ones of \(\displaystyle f(x) = x_i\) where \(\displaystyle x_i\) for i = 1, 2, 3 are the (real) roots of f(x).

.


Hey Balarka,

I really like it you mentioned that to solve for $x$ in \(\displaystyle f(f(x)) = 0\) is equivalent to solve for \(\displaystyle f(x) = x_i\) where \(\displaystyle x_i\) for i = 1, 2, 3 are the (real) roots of f(x).

And if I may use this idea to solve for the problem, I found out the total number of real roots for \(\displaystyle f(f(x)) = 0\) are 7 as well, and this method is so much better than my first approach and thank you, Balarka for showing me something that I don't know!

f(f(x))=0.JPG
Solving for the number of real roots of \(\displaystyle f(x) = x_1=root_1\) means the same thing as finding the number of intersection points between the curve $y=f(x)$ and $y=root_1$, as shown by the green line and we see that in this case we have only one real root for \(\displaystyle f(x) = x_1=root_1\).Solving for the number of real roots of \(\displaystyle f(x) = x_2=root_2\) means the same thing as finding the number of intersection points between the curve $y=f(x)$ and $y=root_2$, as shown by the purple line and we see that in this case we have only one real root for \(\displaystyle f(x) = x_2=root_2\).Solving for the number of real roots of \(\displaystyle f(x) = x_3=root_3\) means the same thing as finding the number of intersection points between the curve $y=f(x)$ and $y=root_3$, as shown by the orange line and we see that in this case we have only one real root for \(\displaystyle f(x) = x_3=root_3\).

So there are a total of 7 intersection points and hence there are 7 real roots for \(\displaystyle f(f(x)) = 0\)!

Thanks!
 

mathbalarka

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MHB Math Helper
Mar 22, 2013
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anemone said:
this method is so much better than my first approach
May I see your approach on this problem?

Balarka
.
 
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anemone

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Feb 14, 2012
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May I see your approach on this problem? Balarka .
My first approach revolves around checking the nature of the curve of the function of $f(f(x))$ from its first and second derivative expression. If we let $f(x)=x^3-3x+1$ and $g(x)=f(f(x))=(f(x))^3-3(f(x))+1$, we find

$f(x)=x^3-3x+1$$g(x)=f(f(x))=(f(x))^3-3(f(x))+1$
$f'(x)=3x^2-3=3(x^2-1)=3(x+1)(x-1)$

$\rightarrow f'(x)=0$ iff $3(x+1)(x-1)=0$ or $x=\pm1$.
$g'(x)=3f(x)^2f'(x)-3f'(x)=3f'(x)((f(x))^2-1)$

$\rightarrow g'(x)=0$ iff $x=\pm1$, $x=\pm \sqrt{3}$, or $x=-2$ or $x=0$.
$f''(x)=6x$$g''(x)=6f(x)(f'(x))^2+3f''(x)((f(x))^2-1)$

I then made another table to determine the nature of the critical points as follows:
$x$-2$-\sqrt{3}$-101$\sqrt{3}$
$f(x)$-1131-11
$f'(x)$960-306
$f''(x)$-12$-6\sqrt{3}$-606$6\sqrt{3}$
$g(x)$3-119-13-1
$g'(x)$000000
$g''(x)$-486216-144540216

And plot a rough sketch to determine the number of real roots of the function of $f(f(x))=0$:

Graph ff((x)).JPG

From the graph we can tell there are 7 real roots of the function of $f(f(x))=0$.
 

mathbalarka

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MHB Math Helper
Mar 22, 2013
573
Hmm... very nice. I like the way you analyzed the behavior of f(f(x)), very well done. Mine may have been a little time consuming but this one is more representative.

PS I particularly like this solution. (Mmm)
 
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anemone

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Feb 14, 2012
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Hmm... very nice. I like the way you analyzed the behavior of f(f(x)), very well done. Mine may have been a little time consuming but this one is more representative.

PS I particularly like this solution. (Mmm)
Thank you for the compliment, Balarka!:eek:

And I must give full credit to Opalg, a very nice mentor here for giving me idea (from another question) to solve the problem!