# Given: f(x) = 1/x-2 g(x) = the square root of x-3 Find f(g(5))

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`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`**

**Sources:**

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

To find `f(g(5)` ), first find g(5)

`g(5) = sqty(5-3) = sqrt 2`

Now substitute this in the function `f(x) = 1/(x - 2)`

`f(sqrt 2) = 1/(sqrt 2 - 2)`

The value of `f(g(5)) = 1/(sqrt 2 - 2)`