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 $(0,1)$.

Let $f(x) = x^5+x^4-1$.
Then we have $f'(x)=5x^4+4x^3=x^3(5x+4)$
There are two turning points, one is at the location when $x=-\frac45$ and the corresponding $f$ value is $-\frac{4^5}{5^5}+\frac{4^4}{5^4}-1 =\frac{-4^5+5(4^4)}{5^5}-1=\frac{4^4-5^5}{5^5}<0$ and the other is at the location when $x$ takes value $0$ with the corresponding $f$ value $-1$.
By observing the sign of the gradient, the graph increases on the interval $(-\infty, -\frac45)$ then reduces on $(-\frac45, 0)$ then increases on $(0,\infty)$. There is a local maximum at $x=-\frac45$ and a local minimum at $x=0$.  Hence all $f$ values on $(-\infty, 0)$ is upper bounded by $f(-\frac45)<0$. there is no negative root.
The graph then increases to positive infinity, hence there is exactly one root which is positive.
