Question

\[f\left( x \right) = \begin{cases}- 4x + 5, & 0 \leq x \leq 1 \\ 2x - 3, & 1 < x \leq 2\end{cases}\] Discuss the applicability of Rolle's theorem for the following function on the indicated intervals ?

Answer 1

The given function is 

\[f\left( x \right) = \begin{cases}{ - 4x + 5, 0 \leq x \leq 1\\2x - 3, 1 < x \leq 2}\end{cases}\]

At x = 0, we have

\[\lim_{x \to 1^-} f\left( x \right) = \lim_{h \to 0} f\left( 1 - h \right) = \lim_{h \to 0} \left[ - 4\left( 1 - h \right) + 5 \right] = 1\]

And

\[\lim_{x \to 1^+} f\left( x \right) = \lim_{h \to 0} f\left( 1 + h \right) = \lim_{h \to 0} \left[ 2\left( 1 + h \right) - 3 \right] = - 1\]

\[\therefore\] \[\lim_{x \to 1^-} f\left( x \right) \neq \lim_{x \to 1^+} f\left( x \right)\]

Thus,  \[f\left( x \right)\]  is discontinuous at  \[x = 1\] 

Hence, Rolle's theorem is not applicable for the given function.