`f(x){{:(2x "," if x lt 0 ),(0"," if 0 le x le 1),(4x "," if x gt 1 ):}` Discuss the continuity

Explanation

`f(x){{:(2x "," if x lt 0 ),(0"," if 0 le x le 1),(4x "," if x gt 1 ):}`
If `x lt lt 1 then f(x) = 2x ` is a polynomial function, which is continuous,
If `0lt x lt 1 then f(x) = 0` is a polynomial function, which is continuous.
If `x gt 1 then f(x) = 4x` is a polynimial function, which is contnuous.
Now at x = 0
f(0) = 0
L.H.L. = ` underset (x to 0^(-))(lim)f(x)`
Let 0 - h - x
`rArr 0 - h to 0 `
`rArr h to 0 `
`=underset (h to 0)(lim) f(1-h)`
`=underset (h to 0)(lim) f(1-h) = 0 `
R.H.L. = ` underset (x to 0^(+))(lim)f(x)`
Let 0 + h = x
`rArr 0 - h to 0 `
`rArr h to 0 `
`=underset (h to 0)(lim) f(1+h)`
`=underset (h to 0)(lim) (0)=0`
`:'` L.H.L. = f(0) = R.H.L.
`:.` f(x) is continuous at x = 0 .
at x = 1
f(1) = 0
R.H.L. = ` underset (x to 0^(+))(lim)f(x)`
Let 1+ h - x
`rArr 1 + h to 1 `
`rArr h to 0`
`=underset (h to 0)(lim) f(1+h)`
`=underset (h to 0)(lim) 4 (1+h)= 4(1+0)=4`
`:'` f(1) `ne` R.H.L.
`:.` f(x) is discontinuous at x = 1 .
Therefore, f(x) is continuous at all points except x= 1