`f(x)={(log_(e)x",",0lt x lt 1),(x^(2)-1 ",",x ge 1):}` and `g(x)={(x+1",",x lt 2),(x^(2)-1 ",",x ge 2):}`.
Then find `g(f(x)).`

Answer

`g(f(x))={(1+"In "x",",0lt x lt 1),(x^(2)",",1 le x lt sqrt(3)),((x^(2)-1)^(2)-1 ",",x ge sqrt(3)):}`

Explanation

`f(x)={(log_(e)x",",0lt x lt 1),(x^(2)-1 ",",x ge 1):}` and `g(x)={(x+1",",x lt 2),(x^(2)-1 ",",x ge 2):}`
`g(f(x))={(f(x)+1",",f(x) lt 2),((f(x))^(2)-1 ",",f(x) ge 2):}`
`={(log_(e)x+1",",log_(e)x lt 2","0lt x lt 1),(x^(2)-1+1",",x^(2)-1 lt 2","x ge 1),((log_(e)x)^(2)-1",",log_(e)x ge 2","0lt x lt 1),((x^(2)-1)^(2)-1 ",",x^(2)-1 ge 2"," x ge 1):}`
`={(log_(e)x+1",", x lt e^(2)","0lt x lt 1),(x^(2)",",-sqrt(3)lt x lt sqrt(3)","x ge 1),((log_(e)x)^(2)-1",",x ge e^(2)","0lt x lt 1),((x^(2)-1)^(2)-1 ",",x le -sqrt(3) " or "x ge sqrt(3)"," x ge 1):}`
`={(log_(e)x+1",",0lt x lt 1),(x^(2)",",1 le x lt sqrt(3)),((x^(2)-1)^(2)-1 ",",x ge sqrt(3)):}`