# Show that $x^5+x+1=0$ has a solution in $\mathbb{R}$

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 ?

• the derivative is always positive, so the function is strictly increasing everywhere. – Will Jagy Jan 19 at 21:56
• 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. – DonAntonio Jan 19 at 22:00
• Oh thanks , but to apply intermediate value theorem shouldn't the domain of the function be connected ? – Pedro Alvarès Jan 19 at 22:03
• @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) – Hagen von Eitzen Jan 19 at 22:16

## 2 Answers

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.

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.