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Let `f: R->R` and `g: R->R` be defined by `f(x)=x^2` and `g(x)=x+1` . Show that `fog!=gofdot`

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Given,` f(x)=x^2` and `g(x)=(x+1) `
`(gof)(x)=g{f(x)}=g(x^2)=(x^2+1).`
`(fog)(x)=f{g(x)}=f(x+1)=(x+1)^2`
Hence, `gofne=fog`
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