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Show $x^5+5x+1=0$ has exactly one solution with $x\in(-1,0)$

To show the equation has only one solution, we have two parts to show: First, to show there exists a solution; second, the solution is uniqueness.

For the first part, we will do the followings:

Let $f(x)=x^5+5x+1$ with $x\in[-1,0]$. When $x=-1$, $f(-1)=-5$. When $x=0$, $f(0)=1$. Now, apply the intermediate value theorem since $f$ is a polynomial and it is continuous on $[-1,0]$, so there exists an $x_0\in (-1,0)$ such that $f(x_0)=0$.

To show the solution is uniqueness. Let $I=(-1,0)$. As $f$ is a polynomial function, $f$ is differentiable on $I$, so $f'(x)=5(x^4+5)$. For all $x\in I$, $f'(x)>0$; thus, $f$ is strictly increasing on $I$. Therefore, by combining results form part one and part two, $f$ has exactly one solution on $I$.


Can someone tell me where I did incorrect? I don't see where is the error. Thanks

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    $\begingroup$ $f'(x)=5x^4+5$, not $5(x^4+5)$. Otherwise this solution looks spot on. $\endgroup$ – vadim123 Nov 20 '15 at 1:55
  • $\begingroup$ Just as vadim123 mentioned, this is the only fault you have. Everything else is totally correct. $\endgroup$ – Mesmerized student Nov 20 '15 at 1:59
  • $\begingroup$ There's only one real solution. There are $4$ more complex solutions. $\endgroup$ – Bernard Nov 20 '15 at 2:55

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