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 $(0,1)$.

If $-1<x<0$ then $x^{5}+x^{4}=x^{4}(1+x) <1$ because $x^{4}$ and $1+x$ both belong to $(0,1)$. If $ x \leq -1$ then $x^{5}+x^{4}=x^{4}(1+x) \leq 0<1$. Hence there is no root with $x <0$. For $x>0$ the function is strictly increasing so it cannot have a second root. 
