Find the number of distinct real roots

[anemone]

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

 

[jacks]

Just asking

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

Is is Right or not

Thanks

 

[mathbalarka]

Quite a nice bit of a problem you've got there. Here's my solution --

Spoiler

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
.

 

[anemone]

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!

 

[anemone]

Spoiler

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).

.

Spoiler

Hey Balarka,

Spoiler

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!

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]

this method is so much better than my first approach

May I see your approach on this problem?

Balarka
.

 

[anemone]

May I see your approach on this problem? Balarka .

Spoiler

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	$-\sqrt{3}$	-1	0	1	$\sqrt{3}$
$f(x)$	-1	1	3	1	-1	1
$f'(x)$	9	6	0	-3	0	6
$f''(x)$	-12	$-6\sqrt{3}$	-6	0	6	$6\sqrt{3}$
$g(x)$	3	-1	19	-1	3	-1
$g'(x)$	0	0	0	0	0	0
$g''(x)$	-486	216	-144	54	0	216

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



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

 

[mathbalarka]

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.

 

[anemone]

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.

Thank you for the compliment, Balarka!

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