Past exam of the mathematics course of the University of Cambridge / 2006 / ia / Paper 1 / 9E / Solution

The intermediate value theorem states that a continuous real-valued function on $[a,b]$ takes every value between $f(a)$ and $f(b)$. To prove it, suppose $f(a)<k<f(b)$ and repeatedly bisect the interval, keeping endpoints whose function values bracket $k$. The resulting nested closed intervals have lengths tending to zero and a common point $c$ by completeness of the real numbers. Both endpoint sequences tend to $c$, so continuity implies $f(c)=k$. Endpoint values are immediate, and reversing the inequalities handles $f(a)>f(b)$.

For the derivative assertion, put $m=(f(b)-f(a))/(b-a)$. The intermediate secant slope from endpoint derivatives can be constructed in three cases. If $m=k$, take $[a^{\prime },b^{\prime }]=[a,b]$. If $m>k$, the functionis continuous on $[a,b]$: differentiability at $a$ supplies the endpoint limit, and differentiability elsewhere implies continuity. Since $g(a)<k<g(b)$, the intermediate value theorem gives $x\in (a,b)$ with $g(x)=k$. Take $[a^{\prime },b^{\prime }]=[a,x]$.

If $m<k$, use insteadIts continuous endpoint extension satisfies $h(a)<k<h(b)$, so there is $y\in (a,b)$ with $h(y)=k$. Take $[a^{\prime },b^{\prime }]=[y,b]$. In every case,The mean value theorem on this subinterval now givesThis proves the Darboux theorem for derivatives without assuming that $f^{\prime }$ is continuous.