15. [1/2 Points]DETAILSPREVIOUS ANSWERSILAFFind the absolute extrema of the function on the closed interval:y = X In(x + 8), [0, 3]minimum(xy)i=0.0maximum(x,y) = (|...

Find the absolute extrema of the function on the closed interval. $$y=x^{2}-8 \ln x,[1,5]$$

For this video. We're going to look at how to find the absolute men and absolute max of a continuous function on a closed interval. So here we have the function y equals three minus the absolute value of t minus three on the closed interval from negative 1 to 5. Now white will three minus the absolute value of t minus three is continuous actually throughout the entire real numbers. So negative infinity to infinity is its domain. Um But we specifically are going to focus on The closed interval from negative 1-5. And definitely if it's continuous over the clothes and the entire interval negative infinity to infinity, it's continuous on the closed interval from negative 1-5. Now to do this, there's a four step process. First we're going to find the critical values of y equal three minus the absolute value of t minus three On the closed interval from negative 1-5. So I'll find the critical values but only work with the ones that are in our specified interval. Then we're going to evaluate the function at the critical values found in step one and we'll write that at the critical values bound in step one. And then step three we evaluate the function at the end points of our clothes dinner fall. So at our left endpoint negative one in our right end 10.5. And then in step forward we compare the output values and steps three and four, the smallest of the output values is the absolute men. The largest of the absolute values of the output values is the absolute max. So let's get started for this function. Finding the critical values of an absolute value function. Um Oftentimes it has us look at the piecewise defined way of writing that absolute value function and then finding the derivative of the pieces. Now for the absolute value function it will not have a derivative at the point that the absolute value turns around. So at that vertex of that absolute value. Looking at this and saying why that occurs another way of writing why equal three minus the absolute value of T minus three is to write it as the piecewise function. Well when you're taking an absolute value of an expression, when the absolute value is taken of zero or a positive number it is its own absolute value. And this T -3 will be zero when he is three and it will be positive when t is greater than three. So I'm gonna write this on the second line, this three minus and then the absolute value of t minus three But in parentheses will still be AT -3. For the T values that are greater than or equal to three because that's when T -3 is positive or zero and the absolute value of a positive or zero number is the number itself. But when you're taking the absolute value of a negative number. So here where T is less than three, something smaller than T might be smaller than three minus three is a negative number. And the absolute value of a negative number is the opposite of that number. So it's three minus The opposite of T -3. Yeah When T. is smaller than three. So looking at that in a simplified way we get this is still the original function. Writing it another way. Why is equal to remove your parentheses and collect like terms you get yeah three minus a negative T. Would be plus T. And then minus a negative negative three is going to give me a negative three. So I just get T. four Cheese smaller than three. And then for distributing the negative sign and collecting like terms I'll get negative t. Plus six. Yeah for the T. S. That are greater than or equal to three. That's the function itself. Now the derivative, well for the tea that are smaller than three, the derivative of tea is just one. And for the teas that are greater than three the derivative of negative T. Is negative one and the derivative of the term six is zero. So I actually get uh derivative of the constant one when the tea is smaller than three and the derivative is equal to the constant negative one when the TSR greater than three. And so since when you're coming in at three from the left and the right on this your limit would not exist as he approached three of your derivative. Um this is not differential at three. Yeah. Yeah. Yeah but just so t equal three was in the domain of my original function. It's in the closed interval that we're considering. Um But it's not differential at that point. Um It's not in the domain of my derivative. So T equals three is a critical value. Yeah. And then when would my derivative equal zero? Well why prime is never zero for this particular thing? It's either one when the teas are smaller than three or its negative one. When the teas are greater than three it's never zero. So T equal three is my critical value. Now in step two it says evaluate my function Y equals three minus t minus three at the critical values founded up one. So why are 3? And let's make that a different color So you can see it better. Y of three is equal to and I'll go back up to the definition that we have with the absolute value intact before we broke it as a piecewise defined function. This is three minus absolute value of take out the tea and put in three And then -3. Well three minus three is zero, absolute value of zero is zero. So why of three Is three subtract 0 which is three. Mhm. Okay next up evaluate the function at the end points of the interval. So at t equal negative one. We're going to evaluate it and then at equal five. So the end points of your interval. Well a T equal negative one. I have y is equal to three minus the absolute value. Take out tea and put a negative one And then -3. So why is equal to three minus the absolute value of negative four? Remember order of operations you've got to take the absolute value of negative four before you subtract. So the absolute value of negative four is four. Three. Subtract four. Let's make these a little bit longer is negative one. Yeah. Mhm. So I have why of negative one input is an output of -1. And then for t equal five I have y is equal to three minus the absolute value of Take out the T. Put in five minus three. So that's why equals three minus five minus three is two. Absolute value to s 23 minus two gives me uh why output of one? So we have y of five Is equal to one. Now we want to compare. So between the numbers Output values of three -1 and one The smallest value is -1. So my absolute man is -1 act a T. Value mm Mhm equal negative one. And so we can also write it as an input of negative one. Gave us an absolute min output of negative one. And then my absolute max is comparing these are absolute max is three right And that happened at T equals three. So we can write an input of three. Okay, gave us a maximum output of three and these are ordered pairs, not intervals. This last thing that I wrote. So just watch your context of the notation um since that's very easy to confuse when it's intermixed within the same work. Right. And that's an example of how to find the absolute extreme A of a continuous function on a closed interval.

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For this problem we are asked to locate the absolute extreme of the function y equals three minus the absolute value of t minus three on the closed interval from negative 1 to 5. The first thing that we want to do is split this up as a piecewise function. So we'll have that this is going to be three minus t plus three When T -3 is less than zero. Or equivalently when T is less than three. And then this will be three minus. Actually no I need to correct myself that up. There would be three plus t minus three on t less than three. And then it will be three minus t plus three four. T greater than three or greater than or equal to three. Which means that we need to then or that we can then differentiate this with respect to T And see that in either case we'll have actually the first case we'll have a derivative of one. And in the second case we'll have a derivative of -1. Which means that we will have a discontinuity at T equals three. So to figure out the absolute max and min We'll need to evaluate our function, adds negative one will give us three minus uh negative one minus three. So I'd be three minus the absolute value of negative four. So that's going to be three minus four or negative one. And then at positive five Which would give us 3- the absolute value of 5 -3. So three minus the absolute value of two which will give us a value of positive too. And then lastly we want to evaluate at three but that would actually I'll correct myself. Um I want to evaluate at three which will give us a value of three minus the absolute value of three minus three. So we get just a value of three which then means that our minimum is going to be the point negative one negative one and our maximum Is going to be the .33.