Question

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

List the critical point(s) of f(x) below in increasing order. If there are any unused boxes, type none.
Critical Points of f(x) occur at x = , , , .
Note: Critical points must be interior points of the domain.

Determine the coordinates of the absolute maximum and minimum values of the function below. If the extrema occurs at more than one point, use the smallest x value. If it does not exist, type none.

Answer 1

I get that the derivative is f'(x) = x/(sqrt(x^2-1). But is there an easy way to get the critical point from that or am I missing something?

Answer 2

Ok, you got the derivative correct. There are a few cases where critical points occur. The definition of a critical point is where f(x) is defined at x = c, but f'(c) = 0 or f'(c) is undefined.

Answer 3

So, first, what is the domain of f(x)?

Answer 4

0 to inf

Answer 5

Wait this is where I screwed up hahah

Answer 6

Not quite. That would be true if \(f(x) = \sqrt{x}\). However, this is a composite function, so find where \(x^2 - 1 >0\)

Answer 7

1 to inf

Answer 8

Wait XD I'm dumb

Answer 9

Sorry, meant to say when \(x^2 - 1 \ge 0\). So, first, find where the "equal" part is true. Factoring, we get \((x+1)(x-1) = 0\). This means that this is true when x = -1, 1.

Answer 10

To find the "greater than" part, we make a number line and test each case.