`f(x) {{:(-2"," if x le -1),(2x"," if -1lt x le 1),(2"," if x gt 1):}`

Explanation

`f(x) {{:(-2"," if x le -1),(2x"," if -1lt x le 1),(2"," if x gt 1):}`
If x `lt` - 1 then f(x) = - 2 is a polynomial function which is contnuous.
If `- 1 lt x lt 1` then f(x) =2x is a polynomial function which is contnuous.
If `x gt 1` then f(x) 2 is a polynomial function which is continuous.
Now at x = -1
L.H.L. = ` underset (x to -1^(-))(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) (-2)=-2`
R.H.L. = ` underset (x to -1^(+))(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) 2( - 1 + h)`
` 2( -1 + 0) = -2`
`:'` L.H.L. = f(-1) R.H.L.
`:.` f(x) is continuous at x = - 1
at x = 1
`f(1)= 2xx 1 = 2`
L.H.L. = ` underset (x to 1^(-))(lim)f(x)`
`=underset (h to 0)(lim) f(1+h)`
`=underset (h to 0)(lim) 2( 1 - h) 2(1-0)=2`
R.H.L. = ` underset (x to 1^(+))(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) 2 = 2`
`:'` L.H.L. = f(1) = R.H.L.
`:.` Therefore, f(x) is continuous for all value of x