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(t) = t³ − 3t², −∞ < t ≤ 3

Accepted Answer

Welcome back, everyone. In this problem, we want to identify the local minimum value of the function P of Z equals Z cubed plus 3 3 Z 2 minus 24 Z in the domain where Z is greater than negative infinity but less than or equal to 2. Now if we're going to identify the local minimum value of the function, OK, then first we need to find the derivative of PF Z, OK. Then we need to set it to 0 to get its critical points, and then we'll need to determine the nature of those critical points. That is if they are minimum or maximum values. So first, let's start by differentiating P of Z with respect to Z. Now we know that PR Z is equal to Z cubed plus 3 Z quad minus 24 Z. So that means the derivative of PF Z. It's going to be equal to 32 + 6 Z minus 24. Now at the critical points. That means the derivative of P of Z will be equal to 0, which then implies that 3 Z 2 plus 6Z minus 24 will be equal to 0. If we divide through by 3, then we'll have Z2 + 2 z minus 8 equal to 0. And now if we factor the quadratic, we have Z + 4 multiplied by Z minus 2 equal to 0. This gives us two critical points. Either Z is equal to -4 or Z is equal to 2. Now that we have our critical points, we need to figure out which one is the maximum and which one is the minimum. Now, since Z equals 2 is the right endpoint of the domain domain, we have to check if it is a local extremal still, OK. And both of them are within the domain, so we know there are candidates for a local extrema. Now to test which one they are, if they are a maximum or minimum, we can use the second derivative test. And by the second derivative test, what it basically tells us, OK, is that if our function is negative, it's a, if our second derivative rather is negative, it's a local maximum, and if our second derivative is positive, it's a local minimum. So let's find the second derivative and plug in both values of Z to determine which it is. No, our first derivative is 3Z2 + 6Z minus 24, so that means the second derivative. Is going to be equal to 6 z + 6. No, that means then that at Z equals -4. Then the second derivative, OK, is gonna be equal to 6 multiplied by -4 + 6. That's -24 + 6, which is -18, and that is less than 0. Therefore, Z equals -4 is a local maximum. Remember we said if the second derivative is negative, it's a local maximum. Next, add Z equals 2, then our second derivative. OK. It is gonna be equal to 6 multiplied by 2 + 6. That's 12 + 6, which is 18 and 18 is greater than 0. Thus, this tells us then that Z equals 2 is a local minimum. Now that we know the Z values for a local maximum and minimum, we can substitute them into our function to find the p values to get the actual maximum, OK, and the minimum. OK. So now then add Z equals -4. If we substitute that into our equation, then P-4 is going to be equal to -4 cubed. + 3 multiplied by -4 minus 24 multiplied by -4. No -4 cubed is -64, 3 multiplied by -4 squared. This should be squared, my apologies, is gonna be 48 and 24 multiplied by -4 is 96. 64 + 48 + 96 equals 80. So P of -4 equals 80. Next, At Z equals 2, P 2 is going to be equal to 2 cubed plus 3 multiplied by 2 square minus 24 multiplied by 2. Which is 8 + 12 minus 48, which equals -28. So what can we conclude? Well, no, this tells us that the local maximum. Is at Z equals -4 with a value of -4 equal to 80. Likewise, the local minimum. Is at Z equals 2 with a value of 2 equals -28. Thanks a lot for watching everyone. I hope this video helped.

Identifying Extrema

In Exercises 41–52:

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

f(t) = t³ − 3t², −∞ < t ≤ 3

Key Concepts

Critical Points

First Derivative Test

Endpoints in a Domain

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