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 $(0,1)$.
 A: Note that $f'(x)=5x^4+4x^3=x^3(5x+4)$. So, $f$ is increasing in $\left(-\infty,-\frac45\right]$ and in $[0,\infty)$ and decreasing in $\left[-\frac45,0\right]$. But $f\left(-\frac45\right)=\frac{256}{3125}<1$. Can you take it from here?
A: 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. 
A: 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!
A: 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.
A: I am not sure if I can make the full prove, but here are some hints.
Let $f(x):=x^5 + x^4 -1$ Its derivative is therefore $f'(x)=5x^4+4x^3 >0 \forall x>0$ Since f(1)=1, f has no zeros in $[1, + \infty ]$ (f will keep growing forever)
Similarly, $f(-1)=-1$ and $\forall x<-1, f'(x)>0$ This means f is always growing in $[-\infty, -1]$
Therefore all zeros for $f$ are in the interval $(-1,1)$, where you already know there is at least one. Now we have to proof there is ONLY one. Let's give it a try while my boss does not come back.
$f'(x)=0$ if, and only if, $x=0$ or $x=-4/5$, this means $f$ is monotonous (it does not twist around, it either goes only up or down) on the intervals $[-1, -4/5]$, $[-4/5, 0]$ and $[0,1]$ 
f(-1)=-1; f(-4/5) <0, so no zeros on $[-1,-4/5]$
f(-4/5)<0; f(0)<0, so no zeros again in $[-4/5, 0]$
We already know there is zero on $[0,1]$, and there cannot be more than one.
Wow! I did it!
