Let `f(x) = {{:(,x +1,"where " x lt 2),(,2x -1,"where" x ge 2):}` then `f'(2)` is equal to

Options

A. 0
B. 1
C. 2
D. done not exist

Answer

D

Explanation

`f'(2)` will exist , if `R 'f (2) = L'f(2)`
Now, `R'f(2) = underset(h to 0)(lim)(f(2+h) - f(2))/(h)`
`= underset(h to 0)(lim)(2(2+h) -1-(4-1))/(h)`
` = underset(h to 0)(lim)(4 + 2h -1-3)/(h) = underset(h to 0)(lim)(2h)/(h) = 2`
and `L'f(2) = underset(h to)(lim)(f(2-h) -f(2))/(-h)`
` = underset(h to0)(lim) (2-h + 1-3)/(-h) = 1`
`therefore " ""Lf(2) ne R'f(2)`
Thus f'(2) does not exists.