Assume $f(x)$ and $g(x)$ are two real-valued functions defined on $[0,1]$, $f(0) > 0$ and $f(1) < 0$. $g(x)$ is continuous on $[0,1]$ and $f(x) + g(x)$ is STRICTLY increasing on $[0,1]$. Prove that there exists $\xi \in [0,1]$ s.t. $f(\xi) = 0$.
I tried to use imitate the proof of zero point theorem, but I cannot find a way to use the increasing of $f(x) + g(x)$. Any ideas about the problem?
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$\begingroup$1. Being monotone, $f+g$ has left (resp. right) limits at each $x\in(0,1]$ (resp. $x\in[0,1)$, hence so does $f$ because $g$ is continuous.
2. $f$ may have jumps discontinuities, that is there may be some points $x\in(0,1)$ with $f(x+)\not=f(x-)$. But because $f+g$ is strictly increasing and $g$ is continuous, any such jump satisfies $f(x+)-f(x-)>0$.
3. To get from $f(0)>0$ to $f(1)<0$, the graph of $f$ must cross the $x$-axis. By 2., the graph can't jump across. A continuous crossing implies a zero! Consider, for example, $\xi:=\sup\{x\in(0,1]: f(x)>0\}$.
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