# Spivak Calculus Chapter 8 "Least Upper Bounds", Problem 3a

The solution manual to Spivak's Calculus has typos sometimes, and other times the solutions are terse. I'd say the solution to the following problem is at least terse.

3a) The proof of Theorem 7-1 showed that there is a smallest $$x$$ in $$[a,b]$$ with $$f(x)=0$$. If there is more than one $$x$$ in $$[a,b]$$ with $$f(x) = 0$$, is there necessarily a second smallest? Show that there is a largest $$x$$ in $$[a,b]$$ with $$f(x)=0$$. (Try to give an easy proof by considering a new function closely related to $$f$$.

Here is Theorem 7-1:

If $$f$$ is continuous on $$[a,b]$$ and $$f(a) <0 < f(b)$$, then there is some number $$x$$ in $$[a,b]$$ such that $$f(x)=0$$.

I am interested in the question posed in bold above. I believe I proved the desired result using a longer proof that mirrored one that was present in the main text. The "easy proof by considering a new function..." presented below is still incomprehensible to me. Here it is verbatim:

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 $$f(x)=0$$.

Can someone provide a better explanation for how this proof works, or perhaps a more verbose version of this proof?

In my approach,let $$x=b+t(a-b)$$ where $$t\in[0,1]$$ so that $$x\in[a,b]$$ also.Then let $$g(t)=f(b+t(a-b))$$,we have $$g(0)=f(b)>0$$ and $$g(1)=f(a)<0.$$Since $$g$$ is continuous on $$[0,1]$$,there is a smallest $$t_{0}$$ with $$g(t_{0})=0.$$Therefore $$x=b+t_{0}(a-b)$$ is the largest $$x$$ with $$f(x)=0.$$