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\noindent {\bf Question}

\noindent Show that $f(x) = |x-2|$ on the interval $[1,4]$
satisfies neither the hypotheses nor the conclusion of the Mean
Value Theorem.


\medskip

\noindent {\bf Answer}

\noindent First, note that $f$ is continuous on $[1,4]$, as it is
the composition of two continuous functions, namely absolute value
and a linear polynomial.  However, $f$ is not differentiable at
$x=2$ (since absolute value is not differentiable at $0$), and so
the hypotheses of the mean value theorem are not satisfied.

\medskip
\noindent To see that $f$ does not satisfy the conclusion of the
mean value theorem, we calculate: $f(4) -f(1) = |4-2| - |1-2| = 2
- 1 = 1$ and $4 - 1 = 3$.  However, for $x > 2$, we have that
$f'(x) =1$ and for $x < 2$ we have that $f'(x) = -1$, and so there
cannot be a point $c$ in $(1,4)$ at which $f'(c) = (f(4) -f(1))/(4
-1) =1/3$.


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