Identifying Extrema

In Exerc...

Question

Identifying Extrema

In Exercises 41–52:

a. Identify the function’s local extreme values in the given domain, and say where they occur.

f(x) = (x + 1)², −∞ < x ≤ 0

Accepted Answer

Welcome back, everyone. In this problem, we want to identify the local extreme values of the function PFX equals X + 4 quad minus 5 in the domain where X is greater than negative infinity, but less than R equal to -3. Now, if we're going to find the local extreme values, first we'll need to differentiate PF X, find its critical values, and then determine their nature. So let's go ahead and do that. Now, given PFX. Is equal to x + 4 quad minus 5, OK, then that means the derivative of PFX is going to be equal to 2 multiplied by X + 4, OK? I know at the critical point. The critical point. Then the derivative of PF X, which equals 2 multiplied by X + 4, is going to be equal to 0. If that's the case, that means X + 4 will be equal to 0, and as such, X will be equal to -4. So X equals -4 is a critical point of our graph. And no, since it's within the given domain, it is a candidate for a local extremal. OK, so let's just make a note of that as well. So since X equals -4. Is in the domain. Uh, open parenthesis, negative infinity, negative 3 close bracket, OK. It is a candidate or it could be. The local Extremal, OK. But no, we need to figure out is this a local minimum, a local maximum, or neither. Well, we can do that by finding the nature of the critical point, OK? So no. The nature of our point to find that we can use the 2nd derivative test. In other words. You can find the second derivative of PF X and if it is positive, the function is concave up, meaning it is a local minimum, and if it is negative, the function is concave done, meaning it's a local maximum. Now in this case, since the derivative of PF X is 2 multiplied by X + 4, then that means the second derivative of PFX is going to be equal to 2. And the second derivative is positive, OK. That means the function is concave up. Function is concave up at X equals -4, meaning it is a local minimum. OK The function is a local minimum or could be a local minimum. Now, to determine the local minimum value, we can evaluate the function at a critical point and the end point, OK? Now, at the critical point, OK. Remember, the local minimum will be the least of value of the function and now at the critical point, we're going to be finding P-4, which equals -4 + 4 quad minus 5, which equals -5. Next, at the end point, we know our endpoint is at -3 because it includes -3. So we'll need to find P-3, which equals -3 + 4 quad, which is 1 minus 53 + 42 minus 5, sorry, which is 1 minus 5, which equals -4. Notice that the least value is when X E equals -4. So thus this tells us that the local minimum. OK. The local minimum is at X equals -4, with a value of P being equal to -5. And there is no local maximum, OK? Because we know that uh at the end point it's -4, OK, but we don't know what the local maximum is, OK, or there's no way for us to find the local maximum. Thanks for watching everyone. I hope this video helps.

Identifying Extrema

In Exercises 41–52:

a. Identify the function’s local extreme values in the given domain, and say where they occur.

f(x) = (x + 1)², −∞ < x ≤ 0

Key Concepts

Local Extrema

Derivative and Critical Points

Domain Considerations

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