# Thread: Extrema on an interval find critical numbers

1. ## Extrema on an interval find critical numbers

I am having some trouble with this problem:

(x-1)/(x+3)

After I have evaluated the problem, no critical numbers exist. But according to the multiple choice answer it states there a critical number.

This is my understanding of critical numbers:

if f'(c)= 0 or f'(c)= undefined and c lies in the domain of the function of f(x)

a critical number exist.

According to everything I done, f'(x) = -3 this number is undefined and also undefined at f(x) so the number is not a critical number.

Could someone verify this for me before I present this to the teacher?

Thanks!

2. Alright so f(x)=(x-1)/(x+3), so to find f '(x) use the quotient rule.

So f '(x)=[(x+3)(1)-(x-1)(1)]/(x+3)^2 = 4/(x+3)^2

so from there you should be able to identify critical points

3. $f(x) = \frac{x-1}{x+3}$

$f '(x) = \frac{(x+3)(1)-(1)(x-1)}{(x+3)^2} = \frac{4}{(x+3)^2}$

Now you should be able to continue. I am not sure how you got your derivative to be -3. Because this is a fraction you should use the quotient rule.