The question asks to use the Rolle's Theorem to prove $f(x)=(x-8)^3$ has only one real zero.
I have already used IVT to prove that there is a zero, but I'm stumped on how to use the Rolle's Theorem to prove there is only one zero, because all other methods I have seen for similar problems employ the fact that the derivative never equals zero, so there is no $a$ and $b$ where $f(a)=f(b)$ using the Rolle's Theorem. But the derivative of this equation does equal zero .. at $x=8$ ..
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I think you are looking at it the wrong way.
Rolle theorem talks about $F(x)$ & makes claims about a root of $F'(x)$.
Here we want to make claims about a root of $f(x)$ , hence $F'(x)=f(x)$ & we will have to figure out what $F(x)$ is.
It is easy to see that $F(x)=\frac{(x-8)^4}{4}+c_1$ will work.
Let $a=8-c_2$ & $b=8+c_2$
When we check $F(a)=F(8-c_2)$ & $F(b)=F(8+c_2)$ , we will get Equality.
$F(a)=F(8-c_2)=\frac{(8-c_2-8)^4}{4}+c_1=c_2^4/4+c_1=Y$
$F(b)=F(8+c_2)=\frac{(8+c_2-8)^4}{4}+c_1=c_2^4/4+c_1=Y$
Hence $F'(x)=f(x)$ must have a root between $a=8-c_2$ & $b=8+c_2$
We then want to show that there can be no other root.
Let there be some $x$ value between $a=8-c_2$ & $b=8+c_2$ , that is , $x=8 \pm c_3$ , which makes $F(x)=F(8 \pm c_3)$ Equalize with the values at $F(a)$ & $F(b)$ , which will then give 2 roots for $F'(x)=f(x)$ within those 2 intervals , by Rolle theorem.
We can easily see that
$F(a) = F(8-c_2) = c_2^4/4+c_1 = Y$
$F(8 \pm c_3) = (8 \pm c_3-8)^4/4+c_1 = c_3^4/4+c_1 = Y$
$F(b) = F(8+c_2) = c_2^4/4+c_1 = Y$
will have no DISTINCT Positive Solution for $c_3$
[[ $ c_2^4/4+c1 = Y = c_3^4/4+c1$ will make $c_2 = c_3$ due to monotonicity within Positive range ]]
Thus there is no other interval & no other root.
Proof with other words , we will have this outline :
There are no three Points where we have Same value for $F(x)$.
Hence , there are no 2 intervals where the ENDPOINTS have Same value for $F(x)$.
Hence , there are no 2 intervals where Rolle theorem can claim $F'(x)=f(x)$ has roots.
Hence , there is only one interval where Rolle theorem can claim $F'(x)=f(x)$ has root.
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Yes, this problem is certainly ridiculous, as comments have pointed out. However, let's make lemonade from lemons and turn this into a challenge: Prove that $f(x)=(x-8)^3$ has a unique root without solving any equations.
We slept through the equation solving part of Sister Agnes’s algebra course. All we remember is how to substitute specific values into functions.
We'll go a step further and prove that every polynomial of the form $f(x)=a(x-8)^n$ for $n\geq 1$, $a\neq 0$ has a unique root at $x=8$.
We prove this by induction on $n$. In the case $n=1$, we have $f’(x)=a\neq 0$, so by Rolle’s theorem we must have $f(c)\neq f(d)$ for all $c\neq d$. In particular, substitution shows $x=8$ is a root, and hence the only root, of $f(x)$.
Suppose we have proved the theorem for $n=k$. Then if $f(x)=a(x-b)^{k+1}$, we have $f'(x)=(k+1)a(x-b)^k$, so by our inductive assumption, $f'(x)$ has a unique root at $x=8$.
We claim this means $f(x)$ has a unique root at $x=8$ as well. Otherwise, (since by substitution $x=8$ is indeed a root), if $r>8$ is another root, then by Rolle’s theorem there is a root for $f’(x)$ strictly between $r$ and $8$, contradicting uniqueness of the root for $f’(x)$.
Arguing similarly for $r<8$, we see $x=8$ is the unique root for $f(x)$.
Assume $f$ has at least two distinct roots. Call these $p_1$ and $p_2$. So $0=f(p_1)$ and $0=f(p_2)$. Then at some point $c$ (where $p_1<c<p_2$) we see $f'(c)=3(c-8)^2$ is zero. This is from Rolle's theorem.
Hence $c=8$. It follows that $p_1<8$ and $p_2>8$. But, for $x<8$ we see $(x-8)^3$ is negative, so $f(p_1)=0$ and $p_1<8$ is absurd. And (if $x>8$) $f(x)>0$ so $f(p_2)=0$ and and $p_2>8$ is absurd.
It follows that $f(x)$ can't have two (or more) real roots. The only other possible choice (that can create a problem for us) is if $f$ has zero real roots. However this is absurd as $f(8)=(8-8)^3$ is zero.