Identifying Extrema

In Exerc...

Question

Identifying Extrema

In Exercises 19–40:

a. Find the open intervals on which the function is increasing and those on which it is decreasing.

b. Identify the function’s local extreme values, if any, saying where they occur.

f(x) = x − 6√(x − 1)

Accepted Answer

Hi, everyone, let's take a look at this practice problem. This problem says for the function HF X, which is equal to X minus the square root of the quantity of X minus 3 in quantity, determine the intervals of increase and decrease and find any local extremes. Now, before we can do anything, we first need to find the domain of our function HFX. And if we look at our function H of X, we see that we have the square root of the quantity of X minus 3. Now, in order for our function H of X to be defined, the quantity underneath the radical needs to be greater than or equal to 0. So we'll need to set X minus 3 to be greater than or equal to 0, and we need to solve this for X. So solving for X by adding 3 to both sides will have X has to be greater than or equal to 3. So that means that the domain of our function. It's going to be the right open interval. From 3 to infinity. So now we have the domain, we now need to look for the intervals over which our function is increasing or decreasing. And to do this, we'll need to find the critical points, and to find those, we'll need to take a derivative of our function H of X. So we'll need to calculate H of X. And that's going to be equal to the derivative with respect to X. Of the quantity of X minus square root of the quantity of X minus 3 in quantity. So taking this derivative, this is going to be equal to. When we take the derivative of the first one, we'll have the derivative of X with respect to X, which is 1. Then we'll have minus the derivative with respect to X of the square root of quantity of X minus 3. And so here we call when when you take the derivative of the square root function, you will get 1 divided by the quantity of 2 multiplied by the square root of the quantity, which is X minus 3 in quantity. And we'll need to also apply our chain rule to this. So we need to multiply this fraction here by the derivative with respect to X of the quantity of X minus 3. And when we take that derivative, that's just going to be to 1. So that means that H of X is going to be equal to 1 minus 1 divided by the quantity of 2 multiplied by the square root of the quantity of X minus 3 in quantity. So now we need to determine our critical points. recall that critical points. Occur whenever our first derivative is equal to 0 or when it's undefined. Now, our function H X here is undefined whenever the denominator in our fraction in the second term is going to be equal to 0. So we're gonna set 2 multiplied by the square root with quantity of X minus 3. In quantity equal to 0 and solved for X. So, dividing both sides by 2 and squaring both sides will have X minus 3 is equal to 0, and then adding 3 to both sides will have X equal to 3. So that is one of our critical points. Now, the other critical point is going to occur whenever H X is equal to 0. So we're going to set 1 minus 1 divided by the quantity of 2 multiplied by the square root of the quantity of X minus 3 in quantity, equal to. 0, and we're going to have to solve this for X. So the first step is to add 1 divided by the quantity of 2 multiplied by the square root of the quantity of X minus 3 in quantity to both sides of the equation. So we'll have 1 is equal to 1 divided by. The quantity of 2 multiplied by the square root of the quantity of X minus 3 in quantity. Now multiplying both sides of the equation by 2 multiplied by the square root of the quantity of X minus 3, we'll get 2 multiplied by the square root of the quantity of X minus 3 in quantity is equal to 1. Dividing both sides by 2 will have the square root of the quantity of X minus 3 is equal to 1 divided by 2. Squaring both sides will have X minus 3 is equal to 1 divided by 4. And then we'll just need to add 3 to both sides. And that will give us X is equal to 13 divided by 4. So here we have two critical points, X equal to 3 and X equal to 13 divided by 4. Now, X equals 3 also happens to be an endpoint of our domain. So what we're going to do is use our critical points here to divide up our domain. So we really are only going to be using the X equals to 13 divided by 4 to find our intervals over which our function is increasing or decreasing. So we're going to break up our domain. And so the first interval that we're going to look at is going to be the open interval from 3 to 13 divided by 4. And in order to determine whether our function, H of X is increasing or decreasing, we'll actually need to look at the sign of our derivative. So recall that if our first derivative is positive, that means our function is increasing over this interval, and if it is negative, that means our function is decreasing. So we're going to choose a point in this interval, so we're going to choose X equal to 3.1. So we're going to calculate each of 3.1. And this is going to be equal to 1 minus 1 divided by the quantity of 2 multiplied by the square root of the quantity of 3.1 minus 3 in quantity. And we can evaluate this expression using our calculator. In doing so, this is going to be approximately equal to minus 0.58. And so here we see that our first derivative is negative, so that means that our function is going to be decreasing over this interval. The next interval that we're going to want to look at is the open interval from 13 divided by 4 to infinity. And again, we'll choose a point in this interval, so we'll just choose X equal to 5. So we'll calculate each of 5, and that's going to be equal to 1 minus 1 divided by the quantity of 2 multiplied by the square root of the quantity of 5 minus 3 in quantity. And again, we can evaluate this expression using our calculator, and this is going to be approximately equal to 0.65. So, here we see that we have a positive quantity for our first derivative, so that means that our function is increasing over this interval. And since our function is changing from decreasing to increasing at X equal to 13 divided by 4, that means that we have a local minimum at that point. So, we're gonna need to evaluate our function at our critical point here. So, we'll look at each of 13 divided by 4. And this is going to be equal to. 13 divided by 4. Minus The square root of the quantity of 13 divided by 4 minus 3 in quantity. And when we evaluate this expression, this is going to be equal to 11 divided by 4. So, as the final answer for our problem, we'll see that our function is increasing. Over the open interval. From 13 divided by 4 to infinity. Our function is gonna be decreasing. On the open interval from 3 to 13 divided by 4. We have a Local minimum. When X is equal to 13 divided by 4, with each of 13 divided by 4. Equal to 11 divided by 4. And if you look for a local Maximum We see that there are actually none. So those are the answers that this problem was looking for. So, I hope that this explanation has been useful, and I'll see you in the next video.

Identifying Extrema

In Exercises 19–40:

a. Find the open intervals on which the function is increasing and those on which it is decreasing.

b. Identify the function’s local extreme values, if any, saying where they occur.

f(x) = x − 6√(x − 1)

Key Concepts

Derivative and Critical Points

Increasing and Decreasing Intervals

Local Extrema

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