Polynomial Twist

Calculus Level 3

True or False?

There exists a non-zero polynomial with real coefficients such that $f(x)$ and $f'(x)$ have the same number of distinct roots.

False True

1 solution

Kushal Bose
Dec 12, 2016

Yes the above statement is true.If $f(x)$ has a repeated roots then after differentiating we get the root of $f(x)$ as a root of $f'(x)$ .

Example: $f(x)=x^3-x^2$ .it has two distinct roots namely $x=0,1$ .Now $f'(x)=3x^2-2x$ .It has roots as $x=0,2/3$ .So,the number of distinct roots is same i.e. $2$

How can you be so sure ?.. can you prove it for a general 'n' degree polynomial ?

Sabhrant Sachan - 4 years, 6 months ago

There's no need to prove it in general. The question is only asking if such a thing is possible, i.e. does at least one such f(x) exist. And by providing one he's proved that yes at least one exists.

Anthony Holm - 4 years, 6 months ago

oh , i misinterpreted the question , thanks for help.

Sabhrant Sachan - 4 years, 6 months ago

Well, even for an n degree polynomial, $x^n$ works similarly.

Calvin Lin Staff - 4 years, 6 months ago

Very easy example: constant zero polynomial $f(x) = 0$ . Both $f'$ and $f$ have same number of distinct zeroes. This is before the question edit.

Michael Huang - 4 years, 6 months ago

but you should consider all the cases. You have completely ignored Rolles theorem

Aditya Bharadwaj - 1 year, 7 months ago

Polynomial Twist

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