Problem 7 Easy Difficulty

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

Absolute maximum at $ 5 $, absolute minimum at $ 2 $, local maximum at $ 3 $, local minima at $ 2 $ and $ 4 $

Answer

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01:19

Fahad P.

02:46

Chris T.

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Video Transcript

We are going to sketch the graph of a function that is continuous and 15. Close interval 1, 5 and has absolute maximum at five Absolute minimum at two. Local maximum at three. And local minimum at two and 4. And we have done that here. So we have this graph and as we can see and this graph we have a local minimum At two. This is the lowest point in the graph When we look close to the number two, but it's also the absolute minimum because if we look at the whole graph is the lowest point. So so what to we have local minimum and an absolute minimum. So we have those two parts. We have also local minimum at four as we as was stated here. Local minima at two and 4. So four. We have the local minimum because if you look close to the number four, the graph of the function has the lowest value at that point at the same happens around two but two, we also have the lowest point in the whole graf and that is the absolute minimum. Now we have a local maximum at three. And that's the case because at three we're at this point and if we look around that point, that is it. Um the higher his value when we look around Or nearly near the value three. So we have their local maximum but it's not a global maximum because the global maximum or the highest point in the graph when we look at the whole domain happens at this point here, which is five as expected Following the statement of sort of maximum five. And there is no There is neither a maximum or minimum at one. And it's important, the graph of this function is a continuous graph. That is we don't have any jumps or points to touch from the trend of the lines, only one piece, draw that his containers function. So we have this example this way of drawing a function with this property continues on 15 That's here to the meaningful and five close interval. And we have an absolute maximum at five here at this point, an absolute minimum at this point two. And we have a local maximum at three here and local minimum add to and four. So that's the that's one solution to the world. Yeah.