Discuss the continous of the following function on its domain, where
`f(x)=x^(2)-4, "for" 0lex le2`
`=2x+3, "for" 2 ltx le4`
`=x^(2)-05, "for" 4 lt x le6.`

Explanation

The domanin of the function is `[0.6].`
For `0le xle2, f (x)=x^(2) -4,` being a polynomial function is continous.
For ` 2 lt x le4, f(x) 2x+3,` being a polynomial function is containous.
For `4 lt x le 6, f(x) =x^(2)-5,` being a polynomial function is continous.
Continuity `x=2`
`f(x) =x^(2)-4,` for `0 le x le 2 therefore f(2) =2^(2)-4=0`
`underset(x to ^(2-))lim f(x)=underset(x to 2)lim(x^(2)-4)=2^(2)-4=0`
`f(x)=2x+3, "for" 2 lt x le 4`
`thereforeunderset(x to 2^(+))limf(x) =underset(x to 2)lim(2x+3) =2(2)+3=7`
`therefore underset(xto2^(+))limf(x)nef(2)=underset(2 to 2^(-))lim f(x)`
`thereofore f` is not continous at `x=2.`
Contuinuity at `x=4`
`f(x) =2x+3,` for `2 lt x le 4 therefore f(4) =2(4)+3=11`
`underset(x to 4^(+))limf(x)=underset(x to 4)lim(x^(2)-5)=4^(2)-5 =11`
`therefore underset(x to ^(4-))limf(x) =underset(x to 4^(+))limf(x)=f(4)`
`therefore f` is continous at `x=4.`
`therefore f` is continous on its domain [0,6] except at the point `x=2,` where it is discontinous.