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My work:

Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be defined by$f(x)=x^5+x+1$

I calculated $f(-2)=-33$ and $f(1)=3$, then $0\in [f(-2),f(1)]$

So by Intermediate value theorem, and knowing that $\mathbb{R}$ is connected, $\exists c\in \mathbb{R}$ such that $f(c)=0$

Therefore the equation given admits a solution. Correct ? And also how could I prove that this solution is unique ?

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    $\begingroup$ the derivative is always positive, so the function is strictly increasing everywhere. $\endgroup$ – Will Jagy Jan 19 at 21:56
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    $\begingroup$ Your solution is basically correct, but $\;\Bbb R\;$ being connected (even path connected) has nothing to do here and you may want to write that polynomials are continuous functions. About uniqueness of the solution read the above comment. $\endgroup$ – DonAntonio Jan 19 at 22:00
  • $\begingroup$ Oh thanks , but to apply intermediate value theorem shouldn't the domain of the function be connected ? $\endgroup$ – Pedro Alvarès Jan 19 at 22:03
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    $\begingroup$ @PedroAlvarès Details may depend on how your specific formulation of the IVT reads. Typical formulations start "Let $f\colon [a,b]\to\Bbb R$ be continuous ..." and while the reason for the theorem to work is the connectedness of $[a,b]$, it is not specifically mentioned as premise (it's just that intervals are connected anyway) $\endgroup$ – Hagen von Eitzen Jan 19 at 22:16
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Yes, your work appears to be correct. To prove this solution is unique, note that

$$f\left(x\right) = x^5 + x + 1$$

means that

$$f'\left(x\right) = 5x^4 + 1$$

which is always $\ge 1$ as $x^4 \ge 0$ for all real $x$. Thus, $f$ is a strictly increasing function.

You could also graph the function to see this.

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For fun:

Assume there is an other zero, i.e.

$x_0,x_1$, where $x_1 >x_0$.

MVT.

$\dfrac{f(x_1)-f(x_0)}{x_1-x_0}=0=f'(a)$, where $a \in (x_0,x_1).$

But $f'(x)=5x^4+1 >0$, contradiction.

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