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Sketch the graph of a function $ f $ that is continuous on $ [1, 5] $ and has the given properties.

Absolute maximum at $ 2 $, absolute minimum at $ 5 $, $ 4 $ is a critical number but there is no local maximum and minimum there.

Graph shown in the video

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Missouri State University

Harvey Mudd College

Baylor University

University of Michigan - Ann Arbor

We're going to sketch the graph of function that is continuous and 15 And has an absolute maximum at to an absolute minimum at five and four is a critical number but there is no local maximum and minimum there. Okay, so the important thing this example is that for is a critical number but there is no local maximum and minimum there. And the thing is that if the function has a relative at that point for example and derivative must not exist because if if it exists then it means you should be zero, there is a judging lunch of your horizontal and in that case we will have a local maximum or minimum. So In order to there is no order that there is no local maximum and minimum there at four but it stills is a critical number. It means the narrative does not exist. For example, if the tunnel line is very good like in this example The change in line and four is a vertical line. So interpretive, it's not a number and uh evidently there is no Maximum nor minimum at four because around for there is to write and to the left there are images are that are bigger and smaller than the image at four. So it's not a law called nara minimum. So with that piece of that part and we can see that at five we have these uh lowest point in the whole graf and two we have the highest point in all the graph and in this case we have drawn and non differentiable A curve at the .2 but that's not a problem. Remember? The only request is the function is continuous on white five and he's the cases there is no discontinuities or jumps at any point and at two had the highest point on the graph of the function. He's also a local maximum, but is has requested the absolute maximum value. And at five we have here the absolute many men. and at four we don't have maximum or minimum and it's a critical point because the derivative does not exist at that point. So this is a solution to the given program.

Universidad Central de Venezuela