Identifying Extrema

In Exerc...

Question

Identifying Extrema

In Exercises 19–40:

b. Identify the function’s local extreme values, if any, saying where they occur.

f(r) = 3r³ + 16r

Accepted Answer

Welcome back, everyone. In this problem, we want to identify any local extreme values for the function PF T equals 40 5 minus 20T cubed. A says the local minimum is P root 3 E equals -24 root 3, and the maximum P negative route 3 is 24 root 3. B says the minimum is P negative route 3 equals 24 root 3, and the maximum is Poff route 3 equals -24 root 3. C the minimum Poff negative route 3 equals -24 route 3, and the maximum p of route 3 equals 24 route 3, and the D at the minimum of negative route 3 equals -24 root 3, while the maximum P of route 3 equals 24. Now, how are we going to identify our local extreme values for our function? Well, to do that, we'll need to find our critical points and then determine the nature of those critical points. Now how can we find the critical points? Well, at our critical points. OK. We know That the first derivative of our function will be equal to 0. So let's differentiate P of T, set it to 0, and solve for T. Now, our first derivative. OK. It's gonna be equal to 20 T4 minus 60T2. So, at the point where our first derivative is equal to 0, OK, then we'll have 20T to the 4th minus 60T 2 equal to 0. If that's the case, And we can factor 20 T squad from the equation to get 20 T squared multiplied by T2 minus 3 to be equal to 0. And if no, if that's the case, then this would imply that 20 T squared would be equal to equal to 0, OK, which means T would be equal to 0. Or t square minus 3 would be equal to 0, which meanst would be equal to plus R minus the square root of 3. In other words, T equals positive root 3 and the negative root 3. This is the t value for our critical point. No, we need to figure out if either of these three values, that is where T equals 0, T equals positive root 3, or T equals negative root 3, if any of these are, uh, maximum or minimum values. Let's use the second derivative test to do that. Recall that the second derivative test basically tells us that if we find a second derivative of our function at the point and it is it is positive, that means the point is a local minimum, but if the second derivative is negative, the point is a local maximum. So now, let's start, let's first find our second derivative and then we can evaluate them at each value of T. Now remember, our first derivative is 20T4 minus 60T2, so that means the second derivative would be 80T cubed minus 120T. Since that's the case And first, starting at. OK, well, let me write that up properly here. By the 2nd derivative test. Can derivative test. OK. Then starting with when T equals 0, then the second derivative. It is gonna be equal to 80 multiplied by 0 cubed minus 120 multiplied by 0, which equals 0. So the second derivative test is inconclusive at T equals 0. It doesn't 0 isn't positive or negative, so we must use another method to determine the nature of this point. Let's use the first derivative test. So instead, applying the first derivative test. Applying the first derivative test, we're going to look at the value of the first derivative on either side of our critical point, OK? And That will help us to figure out uh what if our function is uh going is concave up or concave down and thus a maximum or a minimum. Now, in this case, by applying the first derivative test, OK, we could take one value to the left of T equals 0, OK? For example, if we let T equals -1. OK. Then our first derivative is going to be equal to 20 multiplied by -1 to the 4th, minus 60 multiplied by -1 squad, which is 20 minus 60, which equals -40. So it's less than 0 for T less than 0. So, the function is decreasing before T equals 0, OK? Next, if we take the other side of 4 function, let's say when T equals 1, the other side of 4 critical point rather, then the first derivative will be equal to 20 multiplied by 14th minus 60 multiplied by 1 squad, which equals 20 minus 60, which also equals -40. And that is negative as well. So in this case, before and after the critical point, our function is decreasing. Thus, that tells us that T equals 0. OK, is a point of inflection with no local extreme value, OK? Because it's nature is not changing before, uh, in other words, it's not either decreasing before or increasing after or vice versa, OK. Let's test our next point, OK, that is at T equals root 3. Now at T equals route 3, again, we can apply the second derivative test. The second derivative is going to be equal to 80 multiplied by root 3 cubed, OK, minus 120 multiplied by root 3. If we simplify that, This is going to be equal to 240 root 3, OK, minus 120 root 3, which equals 120 route 3, which is a positive value. Since our second derivative is positive, so since a second derivative is positive. That means T equals route 3 is a local minimum. It corresponds to a local minimum value. And since that's the case, that means that value itself, the value of the function at. T equals route 3 is going to be equal to 4 multiplied by route 3 to the 5th, minus 20 multiplied by route 3 cubed, OK? Which is gonna be equal to, uh, if we simplify here, 36 root 3. Minus 60 root 3, which equals -24 root 3, OK? So T equals route 3 is a local minimum and the minimum value is -24 route 3. Finally, let's test it out for when T equals negative root 3. Again, we plug this into the 2nd derivative. Which is gonna give us 80 multiplied by. Negative root 3 cubed. -120 multiplied by negative root 3. If we do our calculations here similar to the first, then we should get this to be equal to -120 root 3, which is less than 0. Thus, since our second derivative. Is less than 0. T equals negative root 3 is a local maximum. And now that we know it's a local maximum, then we can figure out its value. Because, no, that means if we plug negative route 3 into our function P, then it's gonna be equal to 4 multiplied by negative root 3. Rest to the 5th minus 20 multiplied by negative root 3 cubed. Which when we work that out, it would eventually be equal to 24 root 3. OK. So, T equals negative route 3 is a local maximum at which its value is 24 root 3. Thus our local minimum at P of root 3 equals -24 root 3, and our local maximum is at P-route 3, which equals 24 route 3. If we look back at our answer choices, that tells us then that A is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Identifying Extrema

In Exercises 19–40:

b. Identify the function’s local extreme values, if any, saying where they occur.

f(r) = 3r³ + 16r

Key Concepts

Critical Points

First Derivative Test

Second Derivative Test

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