# Thread: The critical number

1. ## The critical number

1) Find the critical number for $y = sqrt(x^2 - 1)$

I must be doing something wrong with finding the derivative of that function because I end up with $y' = x / sqrt(x^2 - 1)$, so my critical numbers were 0, -1 and 1. But the given answer was 1 and -1. Why was the zero neglected? It made the derivative equal to zero, which should've made it a critical number.

2) $y = (x-1)^(1/3)$

The given answer says that there are no critical numbers and the function is increasing. I don't understand why. The derivative gave me 1 / (3(x-1)^(2/3) so shouldn't 0 be a critical number because it made the function undefined?

2. Originally Posted by Archduke01
1) Find the critical number for $y = sqrt(x^2 - 1)$

I must be doing something wrong with finding the derivative of that function because I end up with $y' = x / sqrt(x^2 - 1)$, so my critical numbers were 0, -1 and 1. But the given answer was 1 and -1. Why was the zero neglected? It made the derivative equal to zero, which should've made it a critical number.
you are correct. 0 is a critical number. maybe it was neglected because of the way the question was asked. are you sure you are not leaving anything out?

2) $y = (x-1)^(1/3)$

The given answer says that there are no critical numbers and the function is increasing. I don't understand why. The derivative gave me 1 / (3(x-1)^(2/3) so shouldn't 0 be a critical number because it made the function undefined?
you mean x = 1 is a critical number, because it makes the denominator zero. yes, that is true.

3. Originally Posted by Jhevon
you are correct. 0 is a critical number. maybe it was neglected because of the way the question was asked. are you sure you are not leaving anything out?
The questions lists a number of functions and asks to find the critical numbers and the open intervals on which the function is increasing/decreasing.

4. Originally Posted by Archduke01
1) Find the critical number for $y = sqrt(x^2 - 1)$

I must be doing something wrong with finding the derivative of that function because I end up with $y' = x / sqrt(x^2 - 1)$, so my critical numbers were 0, -1 and 1. But the given answer was 1 and -1. Why was the zero neglected? It made the derivative equal to zero, which should've made it a critical number.

the function is undefined at x = 0

2) $y = (x-1)^(1/3)$

The given answer says that there are no critical numbers and the function is increasing. I don't understand why. The derivative gave me 1 / (3(x-1)^(2/3) so shouldn't 0 be a critical number because it made the function undefined?

the derivative is undefined at x = 1 , not 0
...

5. Originally Posted by Archduke01
The questions lists a number of functions and asks to find the critical numbers and the open intervals on which the function is increasing/decreasing.
there is still something you are missing. had those been the instructions you were given, then the given answer of 1 and -1 would have been incorrect for the first question still.

Perhaps they asked the question in such a way that the interval you are considering does not include the critical points that were left out

6. Originally Posted by Jhevon
there is still something you are missing. had those been the instructions you were given, then the given answer of 1 and -1 would have been incorrect for the first question still.

Perhaps they asked the question in such a way that the interval you are considering does not include the critical points that were left out
The question doesn't leave any special instructions. Here's what it asked precisely;

"In Excercises 9 - 32, find the critical numbers and the open intervals on which the function is increasing or decreasing. Then use a graphing utility to graph the function."

Afterwards it lists the functions that it wants me to solve. Weird...

7. Originally Posted by Archduke01
The question doesn't leave any special instructions. Here's what it asked precisely;

"In Excercises 9 - 32, find the critical numbers and the open intervals on which the function is increasing or decreasing. Then use a graphing utility to graph the function."

Afterwards it lists the functions that it wants me to solve. Weird...
ok, well skeeter gave the reason why 0 was ignored as a critical point in the first case. the function itself is not defined there. considering the derivative there is pointless. for the second one, x = 1 is a critical point.

8. Originally Posted by Jhevon
ok, well skeeter gave the reason why 0 was ignored as a critical point in the first case. the function itself is not defined there. considering the derivative there is pointless. for the second one, x = 1 is a critical point.
So if the function itself is undefined for a particular number, that means that number can't be used as a critical number?

And as for his second point, yes I know x = 1, that was a typo on my part. The point I was trying to make is that the given answer states that there is no critical number, while you/skeeter are saying that 1 is the crit.

9. Originally Posted by Archduke01
So if the function itself is undefined for a particular number, that means that number can't be used as a critical number?
yes. at least if the function is not just undefined at the point. here, the function is undefined between -1 and 1, a huge gap. the function is not doing anything there, much less increasing/decreasing

And as for his second point, yes I know x = 1, that was a typo on my part. The point I was trying to make is that the given answer states that there is no critical number, while you/skeeter are saying that 1 is the crit.
x = 1 is definitely a critical number. but the function is indeed increasing everywhere else.

10. Originally Posted by Jhevon
but the function is indeed increasing everywhere else.
How are you able to tell?

11. Originally Posted by Archduke01
How are you able to tell?
with test points. we know x = 1 is a critical number (as the derivative is undefined but the function isn't). if we plug in points above and below 1 into the derivative, we see that we get positive answers. hence, the function is increasing.

another way to see this is to notice that the derivative is $\frac 1{3 (\sqrt[3]{x - 1})^2}$, which is always positive. as there is a square in the denominator, and the numerator is a positive number.

12. Originally Posted by Archduke01
How are you able to tell?
because $y' = \frac{1}{3(x-1)^{\frac{2}{3}}} > 0$ for all $x \ne 1$

13. Originally Posted by Jhevon
with test points. we know x = 1 is a critical number (as the derivative is undefined but the function isn't). if we plug in points above and below 1 into the derivative, we see that we get positive answers. hence, the function is increasing.
I see. I just find it strange how the given answer states that the function is increasing even though it says that there is no critical value. Anyway, I'll leave it at that. Thanks so much for you help, broseidon.

14. Originally Posted by Archduke01
I see. I just find it strange how the given answer states that the function is increasing even though it says that there is no critical value. Anyway, I'll leave it at that. Thanks so much for you help, broseidon.
There are many functions that are increasing that do not have critical points. $e^x$ is just one famous one...