If f is a function on [0,1] for which f '(1/2) existsbut f '(1/2) does not belong to [0,1], then f is discontinous at x=1/2. State true or false .Giver reason?
Well as stated there is no problem as $\displaystyle f(x)=100x$ will do as our function.
So lets suppose $\displaystyle f:[0,1] \leftarrow [0,1] $, well then:
$\displaystyle
f(x) = \left\{ {\begin{array}{ll}
0, & {x < 1/4} \\
{2x - 1/2}, & {1/4 \le x < 3/4} \\
1, & {3/4 \le x} \\
\end{array}} \right.
$
RonL