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If $i, j$ are two integers and a function $f$ can be differentiated $i$ times on a open interval. After the $i^{th}$ derivative it will have $j$ roots.

At least & at most how many roots does the original function $f$ have?

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I think at least $0$ root because if

f(x)=2x^2= -1 i=1(means differentiate the function for i times). f^(i)(x) = 0 does not have a root Then F(x) has at least 0 root

How can I find at most how may roots it will have?

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    When you say "solutions", do you mean roots? – 2017-01-05
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    By "solution" do you mean "roots"? But then, take a function like $f(x)=x^2+1$ and $i=1$. $f'(x)$ has a root, namely $x=0$. But $f(x)$ has no roots. – 2017-01-05
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    If $f(x)=x^2+1$ on the interval $[-1,1]$, $f'(x)=0$ has 1 solution but $f(x)=0$ has 0 solutions. – 2017-01-05
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    @pseudoeuclidean yes roots – 2017-01-05

1 Answers 1

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Lower bound: As mentioned in the comments, the original function can have as few as zero roots.

Upper bound:

If $f'$ has $j$ roots, then $f$ has $j$ critical points so it has $\le j+1$ roots. (Try to draw functions with $i$ critical points and see how often you can cross the $x$-axis.) This can be proved rigorously with the mean value theorem.

Repeating this argument for $f^{(i)}$ having $j$ roots shows that $f$ has at most $j+i$ roots.