Critical numbers

Question

Consider the function: √t^2-4 this whole thing is under the square root.

How would I go about figuring out where the function is increasing or decreasing. I've already got the derivative and set it equal to zero, but it has no solution. I usually do the signs diagram for this situation, but I don't know what my interval would be and what numbers I would be able to test to figure out where this function is increasing or decreasing.

Answer 1

Let f(t) be your function. The domain of f is all reals outside the open interval (-2,2). By the chain rule, d/dt(√(t²-4)) = t/√(t²-4); the derivative is positive for t > 0 and negative for t < 0. Hence, the function increases for t ≥ 2 and decreases for t ≤ -2.

Answer 2

The domain of the function is (-inf, -2] U [2,inf), which means -2 and 2 are both in the domain.

When you find the derivative, you find that it is undefined at -2 and 2 (because the square root is in the denominator and cannot divide by zero). So -2 and 2 are critical numbers, so when you draw your number line "sign" diagram -2 and 2 are on it.

Find the sign of the 1st derivative to the left of -2 and to the right of 2.

Check it out here:

desmos.com/calculator/agk7tsgizz

Note the vertical tangent lines x = -2 and x = 2. Those lines are also vertical asymptotes for the 1st derivative.