Sketch the graph of a function continuous on the given interva...

Question

Sketch the graph of a function continuous on the given interval that satisfies the following conditions.

ƒ is continuous on the interval [-4, 4] ; f'(x) = 0 for x = -2, 0, and 3; ƒ has an absolute minimum at x = 3; ƒ has a local minimum at x = -2 ; ƒ has a local maximum at x = 0; ƒ has an absolute maximum at x = -4.

Accepted Answer

Hello there. Today we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Sketch a graph of F that is continuous on the interval, square brackets, 6.6 square bracket. As F of X equals 0 for X equals -5, -3, and 2, absolute minimum at X equals -5, absolute maximum at X equals 6, local maximum at X equals -3, and local minimum at X equals 2. Awesome. So it appears for this particular prom, given all the conditions set to us by the prom itself, we're asked to graph a graph of F. Awesome. So noting that we're trying to sketch a graph of F, that's our final answer we're ultimately trying to solve for. Let us know first off that we're dealing with a closed interval because we have square bracket or inside of square brackets, -6.1, so we have a closed interval -6.1. We're also told that the first derivative of FFX is equal to 0. 4 X equals -5, -3 and 2. We're also told that we have an absolute minimum at X equals -5 and an absolute maximum at X equals 6. And we also have a local maximum at X equals -3 and a local minimum at X equals 2. And we're also told, most importantly, that F is continuous on this close interval of -6.6. So therefore, first off, once again, we should recall and note that since the function is continuous on this closed interval of -6.6, then the function should be smooth without any jumps or breaks across this specific interval. So with that said, therefore we can say that they will that based on what we know that there will be critical points with, let's make a note of this, that F of X, where this prime symbol denotes the first derivative of F of X, it's gonna be equal to 0. And this is going to be at X equals -5, -3 and 2. Awesome, noting that this is a little and symbol. OK. So, with that said, so since the absolute minimum is at X equals -5, the lowest point on the graph will occur at X equals -5, and since there is an absolute maximum at X equals 6, that means the highest point on the graph will occur at X equals 6. So since the local maximum is at X equals -3, there will be a peak at X equals -3, which will mean that this point has a Y value which is larger compared to its immediately adjacent points. And since there's a local minimum at X equals 2, there will be a value at X equals 2, and this will mean that this is at the point at so this will mean that at the specific point it will have a Y value which is smaller compared to its immediately adjacent points. So, with that said, we need to plot these points similar. So we need to plot these points, and we need to note that we do not need to have exact Y values at this point because we're asked to sketch this graph, so we're trying to graph what we know based on our provided information. And we do not know the exact Y values, which is fine, but the important thing in this particular case that we need to focus on is we need to note the position of these points. We need to make sure we're positioning these points properly, like properly, and the position of these points is very important, because we need to consider this in order to figure out which is highest, lowest, and in between in terms of the peaks and falls of our curve. So when we draw our points, we're going to draw our points and then we're going to connect them with a smooth curve, and when we connect them with a smooth curve, we should be able to see the highest and lowest points on this curve, essentially. That is what we're ultimately trying to do. So, let me plot these points and note once again that I'll use a capital I to denote increasing. So if we note that our curve is going to be increasing. At parentheses 53, and at parentheses 26, and let us note that it's going to be decreasing, which I'll say a capital D denotes decreasing at negative infinity 5 in parentheses, and 3.2. So now, considering these pieces of information, we go ahead and plot our points, which I've gone ahead and used some graphing software to plot our points. We can see that we have a point at roughly like X equals -5. We have a point at X equals -3. We have a point at X equals 2, and then we have a point at X equals 6, and if we we plot our four different points and we connect them with a smooth curve, we can see that it increases, and then decreases, and then starts to increase again in an elongated S shape. And it looks like it almost crosses through the origin, but not quite at the origin, but it does cross through the Y axis, and it does intersect the X axis at 3 separate points, and it only intercepts the Y axis one time. Fantastic, and that's it. We have solved for this problem. We have created a sketch. So this problem is very easy to solve for as long as you have an understanding of the graphing methods required to plot your points because you need to be able to recognize. The highs and lows and understand what the what a closed interval is and what an open interval is. And once you have a strong grasp of all the little conceptual things that are associated with graphing a problem like this is very easy to solve. And that's it. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

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