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This question is from Spivak's Calculus (3rd ed) Chapter 8 on Least upper bounds.

Let $f$ be a continuous function on $[a,b]$ with $f(a)<0<f(b)$. Show that there is a largest $x$ in $[a,b]$ with $f(x)=0$.

Before the actual question there is a reference to Theorem 1 which proves there exists a smallest $x$ in $[a,b]$ with $f(x)=0$. This seems to be used in the following answer given in the solutions manual:

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 $g(a)=f(b)>0$ and $g(b)=f(a)<0$. 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 the proof. I don't see why "$b-a+x$ varies between $b$ and $a$ as $x$ varies between $a$ and $b$" and that $g(x)$ satisfies those conditions. Nor can't I see clearly how smallest $y$ for which $g(y)=0$ produces the largest $x$ with $f(x)=0$. I can only guess that it has something to do with the ends points being switched around. Can anyone give a clearer write up of what the solution is trying to do.

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I think you should take $g(x)=f(a+b-x)$. Then you will have $$g(a)=f(b)>0$$ and $$g(b)=f(a)<0.$$ So by the result you mentioned there exists $y$ smallest such that $g(y)=0.$ Then $f(a+b-y)=0$ and $a+b-y$ is the largest such value because if NOT then $g(z)=f(a+b-z)=0$ for some $z$ with $a+b-y<a+b-z$ or $z<y$ and that is a contradiction as $y$ is the smallest number for which $g$ is zero.

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    $\begingroup$ Well gee that makes sense now. I find there are quite a few badly written answers in the official answers book to the spivak text. $\endgroup$
    – helios321
    Jul 29, 2017 at 10:25

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