`f(x)=1/(x-2)` and `g(x)=sqrt(x-3)`

To find `f(g(5))` plug in g(x) wherever x appears in f(x):

`therefore f(g(5))= 1/(sqrt(x-3)-2)`

As x=5 `therefore f(g(5))=1/(sqrt(5-3)-2)`

`therefore = 1/(sqrt2 - 2)`

To rationalize the denominator (ie remove the square root):

`= 1/(sqrt2-2) times (sqrt2+2)/(sqrt2+2)` Note that in essence it remains the same :

`therefore = (sqrt2+2)/(2-4)` The middle term has been eliminated

`therefore f(g(5))=(sqrt2+2)/-2` or `-sqrt2/2+1`

**Ans:**

**f(g(5))= `(sqrt2+2)/-2`**