# Thread: find critical numbers

1. ## find critical numbers

Find all critical numbers for the function f(x) = $\displaystyle \frac {x-1} {x+3}$

2. The critical points are the roots of $\displaystyle f'(x)$.
So, you have to calculate $\displaystyle f'(x)$ and solve the equation $\displaystyle f'(x)=0$.

3. Definitions do differ. In North America it is common to see the following:
A critical number of a function f is a number c in the domain of f at which f (c)=0 or f (c) does not exist.

4. I didn't get it =(

5. Plato gave you a good hint. Find f'(x). Where does f'(x) not exist. What makes the denominator equal to 0?.

6. is it
a) 1
b) 1, -3
c) -3
d) 1, -1
e) None of these

7. Originally Posted by Samantha
$\displaystyle f(x) = \frac {x-1} {x+3}$
Well, what's the derivative of your function?

-Dan

8. Originally Posted by topsquark
Well, what's the derivative of your function?

-Dan
$\displaystyle \frac {1* (x+3) - 1(x-1)} {(x+3)^2} = \frac {x+3 -x+1} {(x+3)^2} = \frac {4} {(x+3)^2}$

9. Originally Posted by Samantha
$\displaystyle \frac {1* (x+3) - 1(x-1)} {(x+3)^2} = \frac {x+3 -x+1} {(x+3)^2} = \frac {4} {(x+3)^2}$
Now, where is this 0 or undefined?

-Dan

10. Originally Posted by Plato
Definitions do differ. In North America it is common to see the following:
A critical number of a function f is a number c in the domain of f at which f (c)=0 or f (c) does not exist.
Since x=-3 is not in the domain of the function and there no c such that f'(c)=0, this function has no critical numbers.

11. Originally Posted by curvature
Since x=-3 is not in the domain of the function and there no c such that f'(c)=0, this function has no critical numbers.
But by definition I believe x = -3 still gets to be a critical number. (As Plato says, it depends on how we define critical numbers.) I agree that this is probably a philosophical point in this case.

-Dan

12. Originally Posted by topsquark
But by definition I believe x = -3 still gets to be a critical number.
I agree with you if we think a critical number is a point where the monotonicity of a function might change.

13. I understand now, thanks =)

14. Originally Posted by Plato
A critical number of a function f is a number c in the domain of f at which f (c)=0 or f (c) does not exist.
Originally Posted by topsquark
But by definition I believe x = -3 still gets to be a critical number agree that this is probably a philosophical point in this case. Dan
I agree with Curvature that the function has no critical numbers. I do not think that it is a philosophical point.

Here is list of the most widely used calculus texts. I have listed them in order of popularity: Stewart; Larson, Hostetler & Edwards; Hughes-Hallett, Gleason, McCallun; Thomas/Finney; Varberg & Purcell; Smith & Minton; and (out of print) Sallas & Hille. In all of these the definition of critical number is exactly as I gave it above. The one exception is found in Thomas/Finney and they require c to be an interior point of the domain. It seems to me this makes a convincing argument that 3 is not a critical number for this function.

15. Originally Posted by Plato
I agree with Curvature that the function has no critical numbers. I do not think that it is a philosophical point.

Here is list of the most widely used calculus texts. I have listed them in order of popularity: Stewart; Larson, Hostetler & Edwards; Hughes-Hallett, Gleason, McCallun; Thomas/Finney; Varberg & Purcell; Smith & Minton; and (out of print) Sallas & Hille. In all of these the definition of critical number is exactly as I gave it above. The one exception is found in Thomas/Finney and they require c to be an interior point of the domain. It seems to me this makes a convincing argument that 3 is not a critical number for this function.
Fair enough. I accept the correction.

-Dan

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