Dgdrxrt

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 $-frac4^55^5+frac4^45^4-1 =frac-4^5+5(4^4)5^5-1=frac4^4-5^55^5<0$ 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.

Note that $f'(x)=5x^4+4x^3=x^3(5x+4)$. So, $f$ is increasing in $left(-infty,-frac45right]$ and in $[0,infty)$ and decreasing in $left[-frac45,0right]$. But $fleft(-frac45right)=frac2563125<1$. Can you take it from here?

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.

Let $xgeq0$ 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:
$$x^5+1-x^4=4cdotfracx^54+1-x^4geq5sqrt[5]left(fracx^54right)^4cdot1-x^4=left(frac5sqrt[5]256-1right)x^4>0$$
and we are done!

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!