Analyzing Functions from Derivatives

Question

Answer the following questions about the functions whose derivatives are given in Exercises 1–14:

b. On what open intervals is f increasing or decreasing?

f′(x) = 1− 4/x², x ≠ 0

Accepted Answer

Hi, everyone, let's take a look at this practice problem. This problem says determine the intervals on which the function H is increasing or decreasing, given its derivative, H of X is equal to 1 minus 25 divided by X2, and X is not equal to 0. So we need to determine the intervals on which our function H is increasing or decreasing. In order to determine these intervals, we'll first need to find our critical points, because those who are going to set the bounds or the various intervals over which our function is increasing or decreasing. Now recall our critical points are located at values of X, where our first derivative is either undefined or equal to zero. So, what the first thing we're going to do is actually rewrite our first derivative. So we'll have H of X. And that's going to be equal to X2 minus 25, and that quantity is going to be divided by X2. So here we just combined everything under a common denominator. And so now we need to determine the values of X for which this function is either undefined or equal to 0. Let's start off with looking at when our first derivative is equal to 0, and that's going to occur whenever our numerator is equal to 0. So we're going to set X2 minus 25 equal to 0 and solve for X. The first thing we'll do is add 25 to both sides of the equation. So we'll have X2d is equal to 25, and taking the square root to both sides, we'll get X is equal to plus or minus 5. So we actually have two solutions here. The other possibility for critical points is when our function F of X is undefined. And that's going to occur whenever our denominator is equal to 0, and our denominator is going to be equal to 0 whenever X is equal to 0. However, X equal to 0 is not in the domain of our functions, since X cannot be equal to 0. So it is not considered a critical point. So really, really just have two critical points to look at. Now we're gonna look at the various intervals that are bounded by our critical points, as well as the location X values, where our function H is also undefined. So, our first interval that we're going to look at is going to be going from minus infinity to -5. And what we need to do is pick a value of X in this interval, so we'll choose X equal to -6, and calculate our first derivatives value. So, we need to look at H. Of minus 6, and this is going to be equal to. The quantity of. 6 in quantity squared, minus 25, and the whole quantity is divided by the quantity of minus 6 in quantity squared. And when we evaluate this expression, this is going to be equal to 11 divided by 36. So here we see our first derivative is going to be positive over this interval, and recall that if the first derivative is positive, that means our function is going to be increasing on that interval. So for our interval going from minus infinity to -5, our function is going to be increasing on that interval. The next interval that we want to look at is going to be going from -5 to 0. That's gonna be the open interval from -5 to 0. And so again, we want to choose a point in this interval, so we're going to choose X equal to -1 and calculate the first derivative. So we'll have H of -1, and that's going to be equal to the quantity of -1 in quantity squared minus 25, and that whole quantity is divided by the quantity of minus 1 in quantity squared. And when we evaluate this expression, this is going to be equal to -24. So, here, our first derivative is negative over this interval, so that means that our function H is going to be decreasing on this interval. Now, the next interval that we want to look at is going to be the open interval from 0 to 5. And again, choosing a point in this interval, we'll choose X equal to 1. We'll calculate each of 1, and that's going to be equal to the quantity of 12 minus 25 in quantity divided by 1 squared. And if I wait in this expression, this is going to be equal to -24. So again, here we see that our first derivative is negative, so that means our function is decreasing on this interval. And then the final interval that we want to look at would be the open interval from 5 to Infinity. And again, we'll choose a point in this interval, so we'll choose X equal to 6, and so we'll have H of 6, and that's going to be equal to the quantity of 62 minus 25 in quantity, divided by 62, and evaluating the expression, this is going to be equal to 11 divided by 36. So, here again, we have a positive first derivative, so that means over this interval that our function H is going to be increasing. So now that we've identified all the intervals over which our function is increasing or decreasing, we can write our final answer. So for our function H. We see that the function is increasing. Over the open interval from minus infinity to -5. Union with the open interval from 5 to infinity. And we see that our function is decreasing. On the open interval from -5 to 0, union with the open interval from 0 to 5. And so those are going to be our final answers for this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

Analyzing Functions from Derivatives

Answer the following questions about the functions whose derivatives are given in Exercises 1–14:

b. On what open intervals is f increasing or decreasing?

f′(x) = 1− 4/x², x ≠ 0

Key Concepts

Derivative and Function Behavior

Critical Points and Intervals

Open Intervals

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