Determining the intervals where f is increasing and where f is decreasing

Question

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

Question: Determine the intervals where f is increasing and where f is decreasing

For this problem when I set the derivative equal to zero, I figured out that this has no solution, but I only figured it out by using an app called mathway, and I was wondering how can I show work to prove this has no solution.

I was also wondering in this type of situation where there are no critical numbers, do I just use the domain of the function to determine where the function is increasing/decreasing.

Answer 1

Answer 2

You're correct that the derivative of this function is never equal to 0; I'll deal with the derivative later on. It's actually possible to solve this problem without using derivatives at all! First, notice that the square root function is only defined if the expression inside the square root is greater than or equal to 0. In this case, we know that

t^2 - 4

will be >= 0 if and only if t is either >= 2 or is <= -2. So our function f(t)=sqrt(t^2 - 4) has a "gap" in the middle: it's not defined between t = -2 and t = 2.

Notice also that our function f(t)=sqrt(t^2 - 4) is an "even" function: what it looks like to the left of the y-axis is a mirror image of what it looks like to the right of the y-axis. Formally, a function g is "even" if g(t) = g(-t). In our case,

f(-t) = sqrt( (-t)^2 - 4) = sqrt(t^2 - 4) = f(t),

so we see that f is even. From here, we just notice that f is increasing on [2,infinity). We can see this simply by noting that t^2 is increases as t increases, meaning t^2 - 4 also increases as t increases, and therefore f(t) = sqrt(t^2 - 4) also increases as t increases. We can go from "t^2 - 4 iincreases with t" to "sqrt(t^2 - 4) increases with t" since the square root function is an increasing function, and applying an increasing function to another increasing function always yields an increasing function as the result.

In case that last paragraph was a little confusing, we can also prove that f is increasing on [2,infinity) by checking that its derivative is positive on [2,infinity). We'll get to derivatives in a minute.

Putting together the facts in bold type above, we see that f is increasing on [2,infinity), not defined on (-2,2), and decreasing on (-infinity,2]. We know it's decreasing on (-infinity,2] since it's an even function and it was increasing on [2,infinity), the "mirror image" of (-infinity,2] across the y-axis.

As for the derivative: Our function is f(t) = [t^2 - 4]^(1/2). Letting g(t) = t^2 - 4, we can write f(t) = g(t)^(1/2). Using the Power Rule and the Chain Rule, we know

f'(t) = 1/2 [g(t)]^(-1/2) * g'(t)

= 1 / [2 sqrt(t^2 - 4)] * 2t

= t / sqrt(t^2 - 4).

Since f'(t) contains the expression sqrt(t^2 - 4), as above we know that f'(t) will not be defined on (-2,2). It also won't be defined at t = 2 or t = -2 since at these points the denominator is zero. But at all other points, namely on (-infinity,-2) and (2,infinity), f'(t) is defined.

Let's examine what happens when t > 2. In this case, the denominator of f'(t) is positive since t > 2 means t^2 > 4. Also, the numerator is positive since t > 2 certainly means t > 0. So if t > 2, then f'(t) > 0. Therefore f is increasing on (2,infinity).

Similarly, if t < -2, then the denominator is still positive since t < -2 means t^2 > 4. The numerator, however, is negative when t < -2. This means if t < -2, then f'(t) < 0. Therefore f is decreasing on (-infinity,-2).

You're correct that the derivative is never equal to 0. The reason it's possible for the derivative to be positive in some places (namely, if t > 2) and negative in others (if t < -2) without ever equaling 0 is because the derivative is not continuous. In our case, it's not continuous simply because it's not defined between -2 and 2.

When there are no critical points of the function f, you can still check where the function is increasing / decreasing by checking if the derivative is positive / negative.