A function is continuous at a point x=c if

(1) `lim_(x->c)` exists

(2) f(c) exists

(3) `f(c)="L"=lim_(x->c)`

We are trying to determine if fg(x) is continuous at x=1:

(1) `lim_(x->1)fg(x)=0` From the left we find `lim_(x->1^-)fg(x)=(3x+5)(x^2-3x+2)==>fg(1)=0`

From the right we determine `lim_(x->1^+)fg(x)=lim_(x->1^+)(sqrt(x)+3)(lnx)=0`

So the limit exists and is 0.

(2) The value of the function fg(1)=0 exists.

(3) The value of the function at x=1 is the same as the limit as x approaches 1.

**Therefore, the function fg(x) is continuous at x=1.**

The graph of f(x) in red, g(x) in blue, and fg(x) in purple:

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