Curve sketching question - Critical points

[hank]

So I have this problem where I'm supposed to sketch the graph:

f(x) = sqrt(4(x^2) - 1)

Ok, so here's what I do...

//Find intercepts
sqrt(4(x^2) - 1) = 0
4(x^2) - 1 = 0
4(x^2) = 1
x^2 = 1/4
x = +/- 1/2

No y intercepts because sqrt(4(0) - 1) doesn't exist.

//Find end behavior
Limits as x->+/- inf are both inf so no horizontal asymptote.

//Find critical points
f'(x) = (1/2)(4(x^2) - 1)^(-1/2)
= 4x / 2sqrt((x^2) - 1)

//Set f'(x) to zero
4x / 2sqrt((x^2) - 1) = 0

Now, here is where I get stuck.

The book says the critical points are (+/-(1/2), 0)

My question is, why isn't +/- 1 critical points as well? If you plug +/- 1 into f'(x), the equation is undefined at that point, and really anywhere in (-1,1).

So why only list +/- (1/2)?

 

[arthur ohlsten]

y=sqrt[4x^2-1] square both sides
y^2=4x^2-1 a parabola, open up, vertex at 0,-1, axis of symmetry x=0

when y=0 x=+/-1/2
Arthur

 

[hank]

I understand that, but what about when you have to find the derivative?

When we go to find the sign of the slope of the tangent line of the curve, the derivative comes out to this:

f'(x) = x / (sqrt(x^2 - 1)

If we set that to zero, any number on [-1, 1] causes the derivative to become undefined.

Why aren't these numbers critical numbers as well?

 

[hank]

bump

 

[hank]

Is it because those numbers aren't within the domain we are interested in?