💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!

# (a) Sketch the graph of a function on $[-1, 2]$ that has an absolute maximum but no absolute minimum. (b) Sketch the graph of a function on $[-1, 2]$ that is discontinuous but has both an absolute maximum and an absolute minimum.

## Graphs shown in the video

Derivatives

Differentiation

Volume

### Discussion

You must be signed in to discuss.

Lectures

Join Bootcamp

### Video Transcript

We have two parts in this problem where a we sketch the graph of a function defined on the closed interval from negative 1-2. That has an absolute maximum, but no absolute minimum in part B. We sketch a graph of the function of the same defined on the same interval. That is this continues but has both an absolute maximum and an absolute minimum. So let's start with a Then we want function that has an absolute maximum but no absolute minimum on a close interval. The function gotta be discontinuous because we know where they stream value theory, that is the function is continuous and the interval disclosed it attains both an absolute maximum value and I've seen absolute minimum value. So fashion cavities continues to have these results. And we have a sketch here. This graph the function is defined on the whole interrupt from there. They want to that is all points. There has an image. We see that we have a discontinuity at this value here. Well, it's already I met that at this value, we have discontinuing because the images over here and the circle, this open circle means the image is not there, but here, up. And for that reason this function has no absolute minimum because the lowest point which should be this open circle here, but it's never attained. So there is no absolute minimum. But the function has an absolute maximum. In fact, on the endpoint, the left hand point negative one. And then the only way to achieve that in a close intervals that the function is discontinuous. So we have here. Okay, uh absolute maximum. Okay, no absolute anymore. F has to be he's continues because the interval where it needs to find his clothes and if the function where continues and the interval closed extreme value theory implies that the function attained its extreme values and the solution will be impossible. But in this case nothing said about the continuities as a function, we deduce we conclude that it has to be discontinuous in. So we draw a discontinuous graph having these properties of attaining an absolute maximum in this case at negative one and know what sort of minimum because the lowest point of the graph is not included in the graph. Yeah. Okay. So we have A and number B. We can we have used exactly the same graph. We have changed the image of this point here. The parade is the image of this number here was here up. But now he's here down. We moved the image only and only doing that. We have a graph of a function to find them negative 12. That has both an absolute maximum here. Yeah. And an absolute anyone here. Yeah. And it is continuous at this point here at this value. This means that the extreme value theorem is not an equivalence. That is if the function is defined and it's continuous is continuous and the close interval then it attains its extreme values. But if the function attains it's a string value doesn't mean automatically that it is continues and this is an example. So it's not an equivalence Is an implication of one only 1 direction. And we can say then that the scruff has a function they find another one to that is discontinues and, yeah, has both absolute maximum and absolute minimum and that's okay, one solution to the problem and given here.

Derivatives

Differentiation

Volume

Lectures

Join Bootcamp