The critical values of a function occur when the first derivative is zero or fails to exist on the domain of the function:

(1) The derivative of a quotient `d/(dx)(f/g)=(gf'-g'f)/(g^2)`

Here `f'(x)=((x^2+5)(1)-(2x)(x-3))/((x^2+5)^2)`

`=(x^2+5-2x^2+6x)/((x^2+5)^2)`

`=-(x^2-6x-5)/((x^2+5)^2)`

(2) `f'(x)` is continuous everywhere, so we need only find the zeros of `f'` :

A fraction is zero if the numerator is zero and the denominator is nonzero. The denominator is greater than zero for all x, so we find when `x^2-6x-5=0`

`x=(6+-sqrt(36-4(1)(-5)))/2`

`=(6+-sqrt(56))/2`

`=3+-sqrt(14)~~6.74,-0.74`

The graph of the original function above:

**So the critical points are** `x=3+-sqrt(14)`

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