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 domain of f is `[0,6]," for "0lexle2,f(x)=x^(2)-4`
`:'` f is a polynomial function, it is continuous in [0,2].
`f(x)=x^(2)-5" for "4ltxle6`
`:'` f is a polynomial function it, is continouns in [4,6].
Thus, f is continuous at every point in `[0,2]uu[2,4]uu[4,6]" for "x=2,f(x)=x^(2)-4`
`:.f(2)=2^(2)-4=4-4=0`
`underset(xto2^(-))limf(x)=underset(xto2^(-))limx^(2)-4=2^(2)-4=0`
`underset(xto2^(+))limf(x)=underset(xto2^(+))lim2x+3=2xx2+3=7`
Thus `underset(xto2^(-))limf(x)neunderset(xto2^(+))limf(x)`
`:.f(x)` is discontinuous at x=2.
Now for x=4f(x)=2x+3
`:.f(4)=2xx4+3=11`
`underset(xto4)limf(x)=underset(xto4^(-))lim2x+3=2 xx4+3=11`
`underset(xto4^(+))limf(x)=underset(xto(4)^+)limx^(2)-5=4^(2)-5`
=16-5=11
`:.underset(xto4^(-))limf(x)=underset(xto4^(+))limf(x)=f(4)`
`:.f(x)` is continouns at x=4.
Hence f(x) is continuous at x=4 but discontinuous at x=2.