In Exercises 1–10, find the extreme values (absolute and local...

Question

In Exercises 1–10, find the extreme values (absolute and local) of the function over its natural domain, and where they occur.

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y = √ 3 + 2𝓍 ―𝓍²

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to identify the extreme values, absolute and local, of the function H of X, which is equal to the square root of the quantity of X2 + 4 X + 3 in quantity over its natural domain. So the first thing we need to do is determine the domain of our function HF X. And HFX involves the square root function, and recall that for the square root function to be defined, we need the quantity underneath the radical to be positive. So that means we need the quantity of X2 + 4 X + 3 to be greater than or equal to 0. Now we can actually factor this quadratic, and it factors into the quantity of X plus 3 in quantity multiplied by the quantity of X + 1 in quantity, and that has to be greater than or equal to 0. So in order for the statement to be true, either both factors need to be positive or both factors need to be negative. And the first factor will change its sign when X is equal to -3, and the second factor will change the sign when X is equal to -1. So if we look at the case when X is less than -3, we see that both factors will actually be negative. So that means that we'll need X to be less than or equal to minus 3 for a domain. Now, for both factors to be positive, if we look at the case when X is greater than -1, both factors will be positive. And so that means we'll need X, B, greater than or equal to minus 1 for the statement to be true. So that means that our domain. is going to be the left open interval from minus infinity to -3. Union With the right open interval from -1 to infinity. So now we just need to determine the extreme values on this domain. And so the first thing we need to do is to calculate H of X to find our critical points. So, H of X is going to be equal to the derivative with respect to X. Of the square root of the quantity of X2 + 4 X. Plus 3 in quantity. So taking this derivative, this is going to be equal to. Recall that when I take the derivative of the square root function, I'll get 1 divided by the quantity of 2 multiplied by the square root of the quantity of X2 + 4 X + 3 in quantity. But here we also have to apply our chain rules, so we need to multiply this by the derivative with respect to x of the quantity of X2 + 4 x + 3. And so when I take that derivative, I get the quantity of 2 X plus 4 in quantity. So simplifying this expression, this is going to be equal to quantity of X plus 2 in quantity, divided by the square root of the quantity of X2 + 4 x + 3 in quantity. So now we need to find the critical points. I recall that the critical points occur whenever our first derivative is equal to zero or undefined. So H X is going to be undefined whenever our denominator is equal to 0, and that's going to occur whenever the quantity of X2 + 4 X + 3 is equal to 0. But we already found those points, and that corresponded to X equal to -3 and X equal to -1. So those are going to be two of our critical points. Now, the other possibility for critical points is when H X is equal to 0, and that's going to occur whenever our numerator is equal to 0. So we'll set X + 2 equal to 0, and solving for X we'll get X is equal to minus 2. And X equal to -2 is not within the domain of our function, so it cannot be our critical point. So we just have two critical points that we need to check. So, we're going to begin by looking at the intervals over which our function is increasing or decreasing. So the first interval that we'll look at is when X is less than -3. And here we just need to really determine the sign of H X over this interval, and the sign of H X is determined entirely by the numerator of a function. Since the denominator is always going to be a positive quantity, we're taking the positive root of the square root function. And so that means if I look at the quantity of X + 2. When X is less than -3, that this quantity is going to be less than 0. And since that is less than 0, that means that H of X. is less than 0. And so that means that our function is decreasing. Over this interval And if we look at the other interval, which is going to be when X is greater than. -1. Again, we'll just look at the numerator over function HXs, and so, When X is greater than -1, the quantity of X + 2 is also going to be greater than 0. And so that means that each of X is going to be greater than 0. So our function is increasing. Over this interval And since our function is decreasing for X less than -3, that means that X equal to -3 is going to be a local minimum. And the same is true for X greater than -1, since our function is increasing over this interval, that means that X equal to -1, since is the lower bound of that interval, is also going to be a local minimum. That means that we have a local minimum. At both X equal to minus 3. And X equal to minus 1. So now we just need to uh evaluate the value of H at each of these two points. So we'll calculate H of minus 3 1st, and that's going to be equal to the square root. Of the quantity of minus 3 quad. Plus 4 multiplied by minus 3. Plus 3 in quantity. And that is going to be equal to 0 when we evaluate that question. And we'll also need to calculate each of minus 1. And that's going to be equal to the square root of the quantity of minus 1 squared. Plus 4 multiplied by minus 1, + 3 in quantity, and by weight in this expression is also equal to 0. So that means that we have a Local minimum value of 0. With H-3 equal to H-1. Which is equal to 0. And if we look for local. Maxima We see that there are none. Our critical points only had. Minimum values. So now that we have the local extrema taken care of, we can look at the absolute extrema. So, for the absolute minimum. It again is going to be equal to 0. And we're gonna have H of minus 3, equal to H of -1, which is equal to 0. Because that is the minimum value that our function H of X will take. And if we look for The absolute Maximum Or a function of H of X, we see that there is none. As X approaches plus or minus infinity, our function H of X is approaching infinity, so it's not bound above, so it has no absolute maximum. So these are the answers that we were actually looking for for this problem, that we have a local minimum with a value of 0 with H of minus 3 equal to H of -1 equal to 0, and a local maximum of none. And we also have an absolute minimum of H-3 is equal to H-1, which is equal to 0, and for absolute maximum, we actually have none as well. 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.

In Exercises 1–10, find the extreme values (absolute and local) of the function over its natural domain, and where they occur.
__________
y = √ 3 + 2𝓍 ―𝓍²

Key Concepts

Extreme Values

Critical Points

Natural Domain

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