Find the critical points of the following functions on the giv...

Question

Find the critical points of the following functions on the given intervals. Identify the absolute maximum and absolute minimum values (if they exist).

ƒ(x) = x³ - 6x² on [-1, 5]

Accepted Answer

Welcome back, everyone, to another video. Determine the critical points of H X equals X to the power of 4 minus 8 X cubed plus 18X2d on the interval from -1 to 5 inclusive. What are the absolute maximum and minimum values. First of all, what we have to do is identify the critical points, so we have to check when H of X is equal to 0 or it does not exist. Now this is a polynomial, so its derivative will also exist for all x values. So we just want to equate the derivative to 0. Now let's differentiate. We're taking the derivative of X to the power of 4 minus 8 x cubed plus 18 x 2d. All that we have to do is apply the power rule, so we're getting 4 X cubed minus 3 multiplied by 8, 24 x 2, plus 18 multiplied by 2, 36 X. And now we can factor it out, right? Essentially we can factor out 4 X. Let's go ahead and do that. Which leaves us with x 2 minus 6 X. Plus 9. From here we can factor out even further because we have a perfect square and this gives us 4 X C X minus 3 squad. So now let's set it to 0. And we have to understand that well, we have a product of two factors, right? In reality, there are 3 factors, but well, we just have one factor squared, so we can just tweed those as 2 factors. So the critical points would be X1 equals 0 because we would have 4 multiplied by 0 multiplied by X minus V2. This gives us 0. And the second solution will be equal to 3 because this is where the second factor is equal to 0, right? So we got our critical points and essentially what we're going to do from here is solve for the absolute minimum and maxima values. So let's go ahead and Perform the analysis. For the critical points, we have one of them being 0, the other one is 3, and we want to check the sign of the derivative or the X values within each interval formed in between, right? So first of all, we take any value that is less than 0. So as we understand, X minus 3 is always non-negative, right? So essentially we can say that X minus 3 is always going to be positive within each of the three intervals on the number line. And now X, well, when it is less than 0, we get a negative factor. When it is between 0 and 3, we get a positive factor. When X is greater than 3, it's also positive. So now the overall product is going to be negative for The interval between negative infinity and zero because we're multiplying a negative number by positive number, then we have a product of 2 positive numbers. So we get a positive derivative between 0 and 3, and then once again from 3 to infinity, we're going to get a positive product. So what does that mean? Well, it basically means that the function is decreasing from negative infinity up to 0, then it starts to increase. From 0 to 3, right? We can extend that line to show that it is increasing from 0 to 3, and then it is still increasing from 3 to infinity, right? So essentially, X equals 3 is a saddle point because the function still keeps increasing. The derivative doesn't change its sign. Now this means that we need to locate. First of all, the minimum value adds 0, and this is the only point of interest, so we want to identify F. Or in this case, H of 0, right? So we're going to take H of 0, and then we want to evaluate the value of the function at the end points. So H at -1 and H at 5. Let's begin with the first one, which is a local minimum point, right? So we're going to take our original function. We're going to take 0 to 4. Minus 8 multiplied by 0 cubed plus 18 multiplied by 0 squared and we're going to get a 0, right? Now, let's identify H at. -1. This gives us -1 to the 4th minus 8 multiplied by -1 cubed plus 18 multiplied by -1 squad, and this gives us 27. And finally, we're taking 5. To the power of 4 minus 8 multiplied by. 5 cubed plus 18 multiplied by 5 squad and this gives us 75 Now, we're looking for the absolute maximum and minimum values, right? So, essentially, we want to identify which of those is the lowest one and the highest one. So the lowest one gives us 0 and the highest one gives us 75, right? So we can conclude that the critical points. Are X equals a set of 0 and 3. Those are the critical x values. Now the absolute maximum would be h max, which means the maximum of the function h. Now we want to specify at which point we get that maximum. Well, our maximum value is 75, we get it at 0.5. And the maximum value is equal to 75. What is the minimum value? H men, the minimum value of H is obtained at 0.0 and it is equal to 0. Those would be our answers to this problem. Thank you for watching.

Find the critical points of the following functions on the given intervals. Identify the absolute maximum and absolute minimum values (if they exist).
ƒ(x) = x³ - 6x² on [-1, 5]

Key Concepts

Critical Points

Absolute Maximum and Minimum

Closed Interval

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