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) = (x − 1)(x + 2)(x − 3)

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says determine the intervals on which our function H is increasing or decreasing, given its derivative H X is equal to the quantity of X + 15 in quantity, multiplied by the quantity of X + 21 in quantity, multiplied by the quantity of X minus 13 in quantity. So we need to determine the intervals on which our function H is increasing or decreasing. Now, in order to determine those intervals, we first need to find the critical points. Recall that critical points occur for the values of X at which our first derivative is either undefined or equal to zero. If you look at our function H X, we see that we basically have a factored form of a polynomial, and recall that polynomials are defined for all values of X. So, therefore, HXx is defined for all real numbers. So, the only way that we can find a critical point is to set our first derivative equal to 0. So, we're gonna set the quantity. Of X plus 15 in quantity, multiplied by the quantity of X plus 21 in quantity, multiplied by the quantity of X minus 13 in quantity equal to 0, and solved for X. Now, in order to solve for X, we're actually going to set each one of these factors equal to 0 and solve for X. So for the first factor, we'll have X plus 15 equal to 0, and therefore X is equal to -15. For the second factor, we'll have X plus 21 equal to 0, and therefore X is going to be equal to -21. And for the last factor, we'll have X minus 13 is equal to 0, and therefore X is equal to 13. So, these critical points are actually going to bound our intervals over which our function is either increasing or decreasing, because we recall that our function has the possibility of changing from increasing to decreasing or decreasing to increasing at our critical points. So, the first interval that we're going to look at is going to be the open interval from minus infinity to -21. And what we want to do is calculate the value of H X at a point in this interval. So, we'll choose X equal to -22. We will calculate H of -22. And this is going to be equal to the quantity of -22 plus 15 in quantity. Multiplied by the quantity of -22 plus 21 in quantity, multiplied by the quantity of -22 minus 13 in quantity. And when we evaluate this this expression, we get -245. And so here we've calculated a negative value for our first derivative, and we call that if a derivative is negative over an interval, that means that our function H is going to be decreasing over that interval. So, our function is actually decreasing. On this interval Now, the next interval that we'll look at is going to be the open interval, going from -21 to -15. And again, we'll choose a point in this interval to calculate our first derivative 4. So, we're going to choose the point X equal to -18. So, we'll calculate H of -18. And that's going to be equal to the quantity of -18 plus 15 in quantity, multiplied by the quantity of -18 plus 21 in quantity, multiplied by the quantity of -18 minus 13 in quantity. And evaluate this expression, we will get. 279. So here we see that we have a positive value for our first derivative on this interval. So since our first derivative is positive on this interval, that means that our function is going to be increasing on this interval. So, the next interval that we're going to look at. It's going to be the open interval from -15 to 13. So again, we're gonna choose a value for X in this interval and calculate our first derivative of. So we're going to choose X equal to 0, so we'll need H of 0, and that's going to be equal to the quantity of 0 plus 15 in quantity, multiplied by the quantity of 0 plus 21 in quantity. Multiplied by the quantity of 0 minus 13 in quantity. And when we evaluate this expression, we will get -4,095. So, here again, we have a negative value for our first derivative on this interval, so that means that our function H is going to be decreasing on this interval. Now, the final interval that we need to look at is going to be the open interval from 13 to infinity. And again, we'll need to choose a point in this interval to calculate our first derivative 4. So we'll choose X equal to 14. We will calculate H of 14, and that's going to be equal to the quantity of 14 plus 15 in quantity, multiplied by the quantity of 14 plus 21 in quantity, multiplied by the quantity of 14 minus 13 in quantity. And when we evaluate this expression, this is going to be equal to 1,015. Since our first derivative is positive on this interval, that means that our function H is going to be increasing. On this interval So to put everything together. We will see that our function is increasing. On the intervals. Of the open interval of -21 to -15. Union with the open interval from 13 to infinity. And then our function is decreasing. On the intervals. Of the open interval from minus infinity to -21, union with the open interval from -15 to 13. So, those are gonna be the 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) = (x − 1)(x + 2)(x − 3)

Key Concepts

Critical Points

Increasing and Decreasing Intervals

Sign Chart Analysis

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