Question

A function is defined by $\left \begin{array}{ll} f(x)= 1 , & \quad for \ \ \frac{-\pi}{2} \ \textless \ x \ \textless \ 0 \\ \hspace{0.75cm}= 1+sinx, & \quad for \ \ 0 \leq x \ \textless \ \frac{\pi}{2} \end{array}$, What can you say about right hand derivatives and left hand derivative at x = 0 ? Is f continuous at x = 0.

Answer 1

f'(0)= lim(h tends to 0+) (f(h) - f(0))/h

= ...........(1 + sin(h) - 1)/h

=............sin(h)/h = 1

this is right hand derivative..

find left hand derivative using the first formula...
left hand derivative will be 0 bcz f(x)=1 is const.
so, it will not be a continuous function.