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I know that $f(0)*f(2) < 0$ so there must be a root between $0$ and $1$ by the intermediate value theorem. But how can I establish that it is unique and simple? All I really know from the IVT is that the root must have an odd multiplicity, since $f(x)$ crosses the x-axis at the root.

symplectomorphic
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Nażysław Zbyłutowicz
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    you simply need to show that $f$ and $f'$ do not have a common root in the unit interval. – dezdichado Mar 12 '17 at 06:44
  • That makes sense, but I have no idea what the root of $f(x)$ is. In my case $f'(0)*f'(1) < 0$ so there still is a root in the interval. How would I show it is different without knowing what the root of $f(x)$ is? – Nażysław Zbyłutowicz Mar 12 '17 at 06:51
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    but you do know what the roots of $f'$ looks like. It could be easier if you argue by contradiction that one of the roots of $f'$ solves $f$ and proceed.. – dezdichado Mar 12 '17 at 06:53
  • Wait, did you do your math correctly? $f(0)=3$ and $f(1)=4$. There's no root between $0$ and $1$. – Bob Jones Mar 12 '17 at 06:54
  • Sorry, that's a typo. I meant $[0,2]$ – Nażysław Zbyłutowicz Mar 12 '17 at 06:56
  • Dezdichado, that makes perfect sense in my example, but I made up this example based on a homework problem, which is slightly different (and I do not want to post, in case my professor is active on here). Would this work in all cases? What would one do if the root of f'(x) is not something that you could easily compute by hand? Then taking more derivatives would cause you to lose some information, wouldn't it? – Nażysław Zbyłutowicz Mar 12 '17 at 07:00
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    @Dradha, if none of $f$ and $f'$ has easily computable roots, then things could get complicated. You would need to find a clever argument depending on the given functions, if they are not simple enough like this. – dezdichado Mar 12 '17 at 07:09
  • Okay, that makes sense. For the problem in my assignment, the method you invoked will be sufficient. Thank you for the help. – Nażysław Zbyłutowicz Mar 12 '17 at 07:14

2 Answers2

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To show that the root is simple, we know that the root is simple if and only if it is also not a root of the derivative.

The derivative of $-x^5+2x+3$ is $-5x^4 + 2$. If both of these share a root $y$, then from the second equation we get $y^4 = \frac 25$. Substituting this in the first equation: $$ -y^5 + 2y+3 = -y^5 + (5y^4)y + 3 = 4y^5+3 = 0 \implies y^5 = \frac{-3}4 $$

Dividing this equation by the earlier one, we get $y = \frac {-15}8$, but it's not even a common root (you can substitute and check), so in fact all roots of the given polynomial are simple, not just the one in the interval $[0,2]$.

Alternatively, you can take the Euclidean gcd of the polynomial and it's derivative to verify this fact.

To see that it's the unique root in the given interval, note that if there were two such roots of the polynomial, then between them would lie a root of the derivative, by Rolle's theorem. So if $a$ and $b$ are roots of the polynomial, with $a<b$, then there is a root of $x^4 = \frac 25$ between $a$ and $b$. Since the real roots of $x^4 = \frac 25$ are less than $1$, it follows that $0\leq a<1$ must hold.

However, if $0 \leq a < 1$,then $a^5 < 1$ , while $2a+3 \geq 3$, so $-a^5 + 2a+3 > 2$ at least, so there can't be a root of the given polynomial in this interval, giving a contradiction.

These two together show that there is a unique simple root of the polynomial $-x^5+2x+3$ in the interval $[0,2]$. Just to mention, that root is approximately $1.42361$.

Sarvesh Ravichandran Iyer
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Let $F(x) = \frac{-x^6}{6} + x^2 + 3x$. Check that $F(x)$ has roots somewhere in the interval $[0, 2]$, so $f$ must have at least one root in $[0, 2]$. Then show that that $F'(x) = f(x)$ is strictly decreasing on the interval $[(\frac{2}{5})^{\frac{1}{4}}, 2]$ ($F''$ is well-defined so this is an easy check), which means $F(x)$ is strictly concave on the interval $[(\frac{2}{5})^{\frac{1}{4}}, 2]$. So $f$ has exactly one root in this interval. For $[0, (\frac{2}{5})^{\frac{1}{4}})$, $f$ is strictly increasing and $f(0) = 3$, so $f$ has no root on this interval. So the root is unique in $[0, 2]$.

Note then that $(\frac{2}{5})^{\frac{1}{4}}$ is a root of $f'$ in $[0, 2]$ but not of $f$.

dasaphro
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