f(x) = ⎧⎪⎨⎪⎩2x, if x<00, if 0≤x≤14x, if x>1
Here, f(x) = ⎧⎪⎨⎪⎩2x, if x<00, if 0≤x≤14x, if x>1
for x <0,f(x)=2x;0<x<, f(x)=0 and x>1,f(x)=4x are polynomial and constant funtion, so it is a continuous in the given interval. So, we have to check the continuity at x =0, 1.
At x = 0 LHL = limx→0− f(x) = limx→0− (2x)
Putting x=0-h as x→0− when h→0
∴limx→0 [2(0-h)] limx→0 (-2h)=−2ltimes0=0,
RHL = limx→0+ f(x) = limx→0+ (0)=0
Also, f(0)=0 ∴LHL=RHL=f(0)
Thus, f(x) is continuous for all values of x.
At x = 1, LHL = limx→1+ f(x) = limx→1+ (0)=0, RHL = limx→1+ f(x) = limx→1+ (4x)
Putting x=1+h as h→1+ when h→0
limh→0[4(1+h)] = (4+4h)=4+4×0=4
∴LHL≠RHL. Thus, f(x) is continuous everywhere except at x=1.