0
$\begingroup$

The following is from Spivak's Calculus.

Let $f$ be a continuous function on $[a,b]$ with $f(a) \lt 0 \lt f(b)$. Then there is a smallest $x$ in $[a,b]$ with $f(x)=0$. We show that there is a largest $x$ in $[a,b]$ with $f(x)=0$ using this fact. Since $b-a+x$ varies between $b$ and $a$ as $x$ varies between $a$ and $b$, the function $g(x)=f(b+a-x)$ satisfies $g(a)=f(b)\gt 0$ and $g(b)=f(a)\lt 0$. So there is a smallest $y$ with $g(y)=0$. Then $x=b-a+y$ is the largest $x$ with $f(x)=0$.

I don't understand why $x=b-a+y$ is the largest $x$ with $f(x)=0$. Shouldn't it just be the $y$ that made $g(y)=0$? Also I don't understand why the text first explains that $b-a+x$ varies between $b$ and $a$ as $x$ varies between $a$ and $b$, then goes on to define $g(x)=f(b+a-x)$. What's the relationship between $b-a+x$ and $b+a-x$ in this reasoning? I'd appreciate it if anyone could explain this question.

$\endgroup$
1
  • $\begingroup$ The nontrivial fact is that there is, say, a smallest $x$ with $f(x)=0$. Given this, it is absolutely obvious that there is also a largest $x$ with $f(x)=0$, and trying to present a formal proof of that along the proposed lines makes it less believable. $\endgroup$ Feb 24, 2015 at 14:24

1 Answer 1

2
$\begingroup$

I think that's a typo in the proof, it should be $b+a-x$ all the time. The expression $b-a+x$ doesn't vary between $b$ and $a$ because for $x=b$ it is $b-a+x = 2b -a \neq a$.

The basic idea of the proof is that you transform the largest root into a smallest root by reverting the order on the domain of the function. That's what the bijective map $[a,b] \to [a,b]$, $x \mapsto b+a -x$ does.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .