Absolute maxima and minima Determine the location and value of...

Question

Absolute maxima and minima Determine the location and value of the absolute extreme values of ƒ on the given interval, if they exist.

ƒ(x) = x³ - 3x² on [-1, 3]

Accepted Answer

Welcome back, everyone. In this problem, we want to find the absolute maximum and minimum values of the function H X equals X cube minus 3 X2 minus 9 X + 5 on the interval open bracket 04 close bracket. He says the absolute minimum is -27 at X equals 3 and the absolute maximum is 5 at X equals 0. B says the minimum is -22 at X equals 3, and the maximum is 10 at X equals -1. C says the minimum is 5 at X equals 0 and the maximum is 21 at X equals 4. And the D says the minimum is -11 at X equals 2, and the maximum is 21 at X equals 4. Now, to find absolute minimum and maximum values of the function, we'll need to find the first derivative of our function. Once we do that, we can use that to help us find a critical point. Now differentiating it of X then. Differentiating H of X. We should get the first derivative to be equal to 3 x 2. 6 X minus 9. Now why is this helpful? Well, at the critical points. At the critical points. The derivative of our function is going to be equal to 0. So in that case, we can set 3 x 2 minus 6 X minus 9 to 0 and solve 4 X to find the X values of the critical points. If we divide through by 3, our equation becomes X2 minus 2X minus 3 equals 0. And if we go ahead and factor. If we factor our terms here, use quadratic factorization, we can rewrite this as X + 1 multiplied by X minus 3 equals 0. And in that case, that means at X equals -1 and at X equals 3, we'll have the critical points for our function. Now, let's evaluate H of X at the critical points and end points, OK? And now when we do that we'll be able to figure out. What our absolute minimum and maximum values will be for our interval, OK? Because at these points, the lowest value will be the absolute minimum, the lowest value of each, while the highest value of H will be the absolute maximum. So let's start with age of. -1. When X E equals -1, H equals -1 cubed, minus 3 multiplied by -1 squad, minus 9 multiplied by -1 + 5. And when we evaluate that H of -1 is 10. Next, 8 of 0 is equal to 0 cubed minus 3 multiplied by 0 squared, minus 9 multiplied by 0, K + 5, which equals 5. Next, at the age of 3. when X equals 3, H equals 3 cubed, minus 3 multiplied by 3 squared, minus 9 multiplied by 3 + 5. Which equals -22. And finally, when X E equals 4, H E equals 4 cubed minus 3 multiplied by 4 squared, minus 9 multiplied by 4 + 5. Which equals -15, OK? No, which of these points is the absolute maximum and minimum. We'll notice that our lowest value is at the point where X equals 3, OK. Thus, this tells us that the absolute minimum is minimum. Let me spell that properly. The absolute minimum is -22 at X equals 3, and notice that our maximum. Is 10 at X equals -1. Thus, the absolute minimum is -22 at X equals 3 and 10 at the absolute maximum is 10 at X equals -1. If we look back at our answer choices, that means answer choice B, is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Absolute maxima and minima Determine the location and value of the absolute extreme values of ƒ on the given interval, if they exist.

ƒ(x) = x³ - 3x² on [-1, 3]

Key Concepts

Critical Points

Endpoints of the Interval

First Derivative Test

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