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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 $f(f(x))=0$.

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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 $f(f(x))=0$.

- Mar 22, 2013

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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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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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- Feb 14, 2012

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Yes, 7 is the correct answer!Just asking

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

Is is Right or not

Thanks

And thanks to you,

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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).

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Hey

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,

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!

- Mar 22, 2013

- 573

May I see your approach on this problem?anemone said:this method is so much better than my first approach

Balarka

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- Feb 14, 2012

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May I see your approach on this problem? Balarka .

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

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

- Mar 22, 2013

- 573

PS I particularly like this solution.

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- Feb 14, 2012

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Thank you for the compliment,

PS I particularly like this solution.

And I must give full credit to