Question

Given f (x) = 2x, x < 0
                 = 0, x ≥ 0 
then f (x) is _______ .

पर्याय

• discontinuous and not differentiable at x = 0

• continuous and differentiable at x = 0

• discontinuous and differentiable at x = 0

• continuous and not differentiable at x = 0

Answer 1

continuous and not differentiable at x = 0

solution:

f (x) = 2x, x < 0

      = 0, x ≥ 0 

${\displaystyle \underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}2x=0}$

${\displaystyle \underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}0=0}$

and f(0) = 0

${\displaystyle \underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}f\left(x\right)=f\left(0\right)}$

Hence, f(x) is continuous at x = 0.
Now we find left hand derivative and right hand derivative of f(0) at x = 0
Right hand derivative at x = 0

i.e ${\displaystyle f\prime \left(0^{+}\right)=\underset{h\to 0^{+}}{lim}\frac{f(0+h)-f\left(0\right)}{h}=\underset{h\to 0^{+}}{lim}\frac{0-0}{h}=0}$

Left hand derivative at x = 0

i.e ${\displaystyle f\prime \left(0^{-}\right)=\underset{h\to 0^{-}}{lim}\frac{f(0+h)-f\left(0\right)}{h}=\underset{h\to 0^{+}}{lim}\frac{h-0}{h}=1}$

${\displaystyle f\prime \left(0^{+}\right)\ne f\prime \left(0^{-}\right)}$ Hence, f(x) is not differentiable at x = 0.