All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math > Calculus
composite function (Posted on 2016-11-29) Difficulty: 2 of 5
It is given that f(x) = x^3 - 3x + 1 .

Find the number of real roots of f(f(x))

No Solution Yet Submitted by Danish Ahmed Khan    
Rating: 3.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution | Comment 1 of 2
f(x) has three real roots at x=-1.8794, 0.3473, and 1.5321

f(f(x)) can have up to three real roots corresponding to each root of f(x). Then we want to know for which values v does the equation f(x)=v have three real roots.  This will occur when the line y=v intersects the curve y=f(x) three times, specifically when the line is between the two relative extrema.

f'(x) = 3x^2 - 3.  f'(x)=0 has roots x=1 and x=-1.  Then the relative extrema of f(x) are (1,-1) and (-1,3).  The y-coordiantes denote the range we seek: f(x)=v will have three real roots when -1<v<3.

Two of the three roots are in that interval, the other is outside.  Then the total number of real roots of f(f(x)) is then 2*3 + 1*1 = 7 real roots.

  Posted by Brian Smith on 2016-11-29 11:05:23
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (0)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information