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 = √𝓍² ― 1

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says determine the extreme values, both absolute and local, of the function G of X, which is equal to the square root of the quantity of 4 minus X squad in 20, over its natural domain and where they occur. So this problem wants us to find the extreme values for our function over its natural domain. The first thing we need to do is determine the domain of our function GFX. And if we look at our function G of X, we have the square root of a quantity, and in order for our function G of X to be defined, we need the quantity underneath the square root to be greater than or equal to 0. So we need 4 minus X2 to be greater than or equal to 0, and this is true whenever x2 is greater than or equal to 4. So that means that X can be between -2 and 2. So that means our domain is going to be the close interval, going from -2 to 2. So next we need to determine our critical points for our function. For that, we'll need to take a derivative of a function G of X with respect to X. We'll calculate G of X. Which is equal to the derivative with respect to X. Of the quantity of 4 minus x squared in quantity raised to the 12 power. So here we've just rewritten the square root, as raising that quantity to the 1 half power, and this will allow us to take the derivative a little bit easier. So, what we want to do is apply our power rule and chain rule to this function to takes derivative. So, what we're going to do is use U substitution for our chain rules. So, we'll say U equal to 4 minus X squared. So that means we can rewrite our function G of U as being equal to. You raise to the one half power. Did not need to determine the derivative of U with respect to X. And when I take the derivative of you with respect to X, this is going to be equal to -2 X. So here we just use our power rule. And we also need to calculate the derivative of G with respect to you. And this is going to be equal to 1/2 multiplied by U raised to the minus 12 power. So again, here we've used our power rule. So now we want to do is use this information to calculate our G of X. So we call for the chain rule that we're going to have the derivative of G with respect to U, which is going to be 1/2 multiplied by the quantity of 4 minus X2 in quantity raised to the minus 12 power. And we need to multiply that by the derivative of U with respect to X, but we need to multiply this by the quantity of minus 2 X. And we can actually simplify this expression, so this is going to be equal to minus x divided by the square root of the quantity of 4 minus X squad in quantity. So here we've just rewritten the quantity of 4 minus X squared in quantity raised to the minus 12 power as 1 divided by the square root of the quantity of 4 minus X squared in quantity. So now, in order to find our critical points, recall that our critical points are the X values at which our first derivative is either equal to zero or undefined. Now, our first derivative here is equal to 0, when our numerator here is equal to 0, since our denominator here is going to be um a positive quantity if it's not equal to 0. So, for our numerator to be equal to 0, we just have minus X equal to 0, so that means X has to be equal to 0. So that's gonna be one of our critical points. And the other condition to find a critical point is when our first derivative is undefined, and our first derivative is undefined when our denominator is equal to 0. So that means I'm gonna have the quantity of 4 minus X squared equal to 0. That means that X can be equal to plus or minus 2. So we actually have 3 critical points here. Now, the X equal to plus or minus 2, those are also the end points of our domain. So, we'll actually analyze those separately, since they can only be absolute maxima or minima. And so we're gonna focus on the X equal to 0 here to see if it's a local minimum or a local maximum. So to do that, we also need to look to see how the sign of our first derivative changes about X equal to 0. Now, if we look at the case when X is less than equal 0, sorry. If we look at the sign of G of X, If X is negative, then that means that our numerator and our function G of X is going to be positive since it's multiplied by a negative sign, and our denominator is always going to be positive since we're taking the positive square root. So that means that G of X is always going to be greater than 0. And if we look at X greater than 0, G X is going to be less than 0, since our numerator is always going to be negative, so we'll have a positive value for X, multiplied by the negative sign out front, which gives us a negative value, divided by a positive number, which gives us overall a negative value for our first derivative. So, what that means is that since the first derivative is changing from a positive value to a negative value, we actually have a local maximum at X equal to 0. And that's because our function is going from increasing to decreasing. So we have a local maximum. At X equal to 0. And we need to determine the value of our function G of X at X equal to 0. So doing so, we'll have G of 0, and that's going to be equal to. The square root Of the quantity of 4 minus 0 squared in quantity, which is equal to 2. So we have a local maximum at X equal to 0, with G of 0 equal to 2. So now we need to look for absolute maximum and minimum, and for this we need to also include our end points. So we'll evaluate our function G of X at our end points, and so we'll have G minus 2. And that's gonna be equal to the square root of the quantity of 4 minus the quantity of -2 in quantity squared in quantity, which is going to be equal to 0. And if we look at G of 2, this is going to be equal to the square root of the quantity of 4 minus 2 squared in quantity, which is also equal to 0. So we can now determine our absolute maxima and minima. And for the absolute maximum. That's going to occur. At X equal to 0 with G of 0 equal to 2. Since that is the largest value that our function G of X obtains, and we actually have Absolute. Minima At X equal to. Plus or minus 2. With G 2 equal to G minus 2, which is equal to 0. So those are the results for our absolute maximum and our absolute minima. And so, just to recap of what we have, we found that we had a local maximum at X equal to 0, with G of 0 equal to 2, and we actually have no local minimum. We also found that the absolute maximum occurred at X equal to 0 with G 0 equal to 2, and we have absolute minima at X equal to plus or minus 2 with G2 equal to G minus 2, which is equal to 0. So those are the answers to this problem. 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 = √𝓍² ― 1

Key Concepts

Extreme Values

Critical Points

Natural Domain

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