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Explanation

Put `x=0, y=0`, then g(0)=0
and `g^(')(0)=lim_(hto0)(g(h))/(h)=2`
Now, `g^(')(x) = lim_(hto0)(g(x+h)-g(x))/(h)`
`lim_(hto0)(e^(h)g(x)+e^(x)g(h)-g(x))/(h)`
`=g(x)lim_(hto0)(e^(h)-1)/(h) + e^(x)lim_(hto0)(g(h))/(h)`
`=g(x)+2e^(x)`
Let `g(x)=y`. Then `(dy)/(dx) = y+2e^(x)`
or `e^(-x)(dy)/(dx)-ye^(-x)=2`
or `d/(dx)(ye^(-))=2`
or `ye^(-x)=2x+c`
Given, if `x=0, y=0`, then c=0
Now, `(dy)/(dx)=2[e^(x)+xe^(x)]=0` or `x=-1`
Thus, minima is at `x=-1`
Thus, range is `[-2/e, infty]`