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) = 12t − t³, −3 ≤ t < ∞

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to find the local extrema of the function H of Y, which is equal to YQ -27Y in the domain of minus infinity is less than Y, which is less than or equal to 5. So this problem asks us to find the local extrema of our function H of Y. To do this, we first need to find the critical points. And to do that, we need to take a derivative of our function H of Y with respect to Y. So we need to calculate H of Y, and this will be equal to the derivative with respect toy of the quantity of Y cubed minus 27 Y in quantity. So to take this derivative, I'm gonna need to use our power rule as well as our constant multiple rule. So taking this derivative, this is going to be equal to 3 Y2 minus 27. Now recall that for critical points, those are going to be the Y values, at which our function H Y is either undefined or equal to 0. Now, if you look at H Y, we see that we have a polynomial, and polynomials are defined for all values of Y. That means that our first derivative here is defined for all values of Y. So the only possible critical points are going to come from setting our first derivative here equal to zero. So we're going to set 3 Y squad. Minus 27 equal to 0, and solve for Y. So first step is to simplify our expression by dividing both sides of the equation by 3, and so we'll have Y squared, minus 9 is equal to 0. We can factor the left hand side of this equation, so we'll have the quantity of Y plus 3 in quantity multiplied by the quantity of Y minus 3 in quantity, and that's will be equal to 0. So now we can set each factor equal to 0 and solve for Y. So for the first factor we'll have Y plus 3 is equal to 0, and therefore, Y is equal to -3. And for the second factor we'll have Y minus 3 equal to 0, and solving for Y we'll get Y is equal to 3. So we have two points to look at, Y equal to -3 and Y equal to 3, and both of those points fall within the domain of our function. So we need to determine whether we have a local minimum or a local maximum at each of these points. And to do this, we're going to use the second derivative test. So, we need to calculate the second derivative, so we need to calculate H of Y, and that's going to be equal to the derivative with respect toy of the quantity of 3 Y2 minus 27 in quantity. And so to take this derivative, we're going to need to use our constant multiple rule as well as our constant rule for derivatives. So taking this derivative, this is going to be equal to 6 Y. So now we need to calculate the second derivative at each of our critical points. So, the first one we'll look at is when Y is equal to -3, so we'll have H of -3, and that's going to be equal to 6, multiplied by -3, which is equal to -18. And so recall that if our second derivative is negative at a critical point, that means that we're going to have a local maximum. So we have a local maximum. At Y is equal to -3. Next, we'll look at H of 3, and that's going to be equal to 6, multiplied by 3, which is equal to 18. So here our second derivative is positive, so that means that we have a local minimum. At Y is equal to 3. So now we've determined whether we have a local minimum or maximum, we just need to calculate the value of H of Y at our two critical points. So, looking at each of minus 3. This is going to be equal to the quantity of -3 in quantity cubed, minus 27 multiplied by -3. And when we evaluate this expression, this is equal to 54. We also need to calculate each of 3, and that's going to be equal to 3 cubed. -27 multiplied by 3. And when we evaluate this expression, this is going to be equal to minus 54. So the final answer for this problem is that we'll have a Local minimum. When Y is equal to 3. And we'll have each of 3. Equal to minus 54. And we'll also have a local maximum. At Y is equal to -3. And then we'll have H of minus 3. Equal to 54. So those are the answers to this problem that we were looking for. So I hope that this explanation has been useful, and I'll see you in the next video.

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) = 12t − t³, −3 ≤ t < ∞

Key Concepts

Critical Points

First Derivative Test

Domain Considerations

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