<s>[INST] Show that the equation $x^5+x^4=1$ has a unique solution. Show that the equation $x^5 + x^4 = 1 $ has a unique solution.
My Attempt:
Let $f(x)=x^5 + x^4 -1 $
Since $f(0)=-1$ and $f(1)=1$, by the continuity of the function $f(x)$ has at least one solution in $(0,1)$.
 [/INST] Let $x\geq0$ and $f(x)=x^5+x^4.$
Thus, since $f$ increases, continuous, $f(0)<1$ and $f(1)>1$, we obtain that our equation has an unique root.
Let $x<0$.
Thus, $-x>0$.
We'll prove that the equation $-x^5+x^4=1$ has no roots for $x>0$.
Indeed, by AM-GM we obtain:
$$x^5+1-x^4=4\cdot\frac{x^5}{4}+1-x^4\geq5\sqrt[5]{\left(\frac{x^5}{4}\right)^4\cdot1}-x^4=\left(\frac{5}{\sqrt[5]{256}}-1\right)x^4>0$$
and we are done!
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