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) = (4x³/3) + 5x² - 6x on [0,5]

Accepted Answer

Welcome back, everyone. In this problem, we want to find the absolute maximum and minimum values of H on the interval open bracket 1 foreclosed bracket. H of X is equal to 2 X minus 15 X2 plus 36 X minus 20. A says the absolute maximum is 8 at X equals 3 and the absolute minimum is 7 at X equals 2. B says the absolute maximum is 8 at X equals 2 and the minimum is 7 at X equals 3. C says the maximum is 12 at X equals 4 and the minimum is 3 at X equals 1. And B says the maximum is 12 at X equals 1 and that the minimum is 3 at X equals 4. Now, before we can find the absolute maximum and minimum values, it would help for us to find the derivative of H of X. So differentiating H of X and we'll get to why we differentiate it in a second. Then the derivative of H of X. Is equal to 6 X squared. Minus 30 X plus 36 Now what do we want the derivative for? Well, recall that at the critical points, OK, at the critical points. Of our function, the first derivative is equal to 0. So if we can set our first derivatives to 0, we'll find the X values of the critical points and then we can evaluate our function at the critical points and endpoints to find the maximum and minimum absolute values. So, let's go ahead and do that. Here, if we set it to 0, that means 6 X2 minus 30 X plus 36 equals 0. If we divide through our equation by 6, we get X2 minus 5 X plus 6 to be equal to 0. And know if we quadratically factorize this expression, we get X minus 2 multiplied by X minus 3 to be equal to 0, which means that the critical points are at X equals 2 and X equals 3. OK. Let's highlight that in red here. Very important. So these are our critical, the X values of our critical points. Now that we have those values and the X values of our endpoints because we're looking at the interval from 1 to 4 inclusive, then that means, OK, that evaluating, we can evaluate H of X. At those points, OK. Which means we're going to look for H of one. HR 2, OK. Age of 3 and age of 4. Now let's start with H of 1. If we plug 1 into our equation, it's going to be 2 multiplied by 1 cubed minus 15 multiplied by 1 squared, plus 36 multiplied by 1 minus 20. I know when we evaluate that, OK, we should get this to be equal to 3. Likewise, we can go ahead and evaluate the rest. I won't write them all out, but when X equals 2, H equals 8, when X is 3, H is 7, and when X is 4, H is 12. What do we notice here? Well, by comparing the values, we can tell that the absolute maximum value is at the age of 4, OK? Where H equals 12. On the other hand, the absolute minimum value is at X equals 1, where H equals 3. Therefore, we can conclude that the absolute value of H. Yeah, absolute value, uh sorry, absolute maximum rather. Absolute maximum of H is 12 at X equals 4, OK. And the absolute minimum. The absolute minimum. Is that here. 3, OK. A X equals 1. If we look back at our answer choices, that tells us that answer choice C 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) = (4x³/3) + 5x² - 6x on [0,5]

Key Concepts

Critical Points

First Derivative Test

Endpoints of the Interval

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