Identifying Critical Numbers for Square Root and Rational Functions

[tjohn101]

Homework Statement

Find all critical numbers of:
g(x)= sqrt(x²-4)
and
f(x)= (1)/(x²-9)

Homework Equations

n/a

The Attempt at a Solution

1) sqrt(x²-4)
simplified to x(x-2)(x+2)^-1/2 (please check)
and got zeroes as x=0, x=2, x=-2
and I got confused because if you do g(2) and g(-2) it equals zero. For some reason I can't tell if they are defined or undefined. x=0 works, so that is a critical number. The other two are throwing me off.

2) (1)/(x²-9)
zeroes were x=-3, x=3, x=0. Plugging 3 and -3 back into f(x) gave me undefined, so I'm pretty sure 0 is the only critical number.

If you could please check the second and help with the first that would be great. Thank you

 

[statdad]

About

"1) sqrt(x2-4)
simplified to x(x-2)(x+2)-1/2"

- is that supposed to be the derivative of $$\sqrt{\,x^2 -4}$$? if so, it isn't correct.

what is your definition of a critical value? (writing it out can help you see the appropriate path)

 

[tjohn101]

I'm honestly not sure what I did there... I re-did my derivative and found

x/((x²-4)^1/2)

which would go to x/(((x+2)(x-2))^1/2)

and give zeroes of -2,2,0. -2 and 2 would make the denominator 0 and be undefined, leaving 0 as the only CP. Sound right? Sorry if any errors I did that really quick.

 

[statdad]

yes, the derivative is

$$\frac{x}{\sqrt{\, x^2 - 4}}$$

and this is zero or undefined for $0 \text{ and } \pm 2$.

Again, what is your definition of a critical number? (same as critical value)

 

[tjohn101]

Definition is:
x=c is a critical number for f(x) if f(c) is defined and f'(c)=0 or f'(c) is undefined.

So that means 2 and -2 WOULD be CPs?

 

[statdad]

Definition is:
x=c is a critical number for f(x) if f(c) is defined and f'(c)=0 or f'(c) is undefined.

So that means 2 and -2 WOULD be CPs?

yes.

 

[tjohn101]

And 0 would not be one because it is undefined in f(x) (sqrt of a negative makes it undefined), correct?

And for the second problem, zero was the only one defined in f(x). Therefore it IS indeed a CP.