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Numerade Educator

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Problem 28 Easy Difficulty

Sketch the graph of $ f $ by hand and use your sketch to find the absolute and local maximum and minimum values of $ f $. (Use the graphs and transformations of Section 1.2 and 1.3).

$ f(x) = \left\{
\begin{array}{ll}
2x + 1 & \mbox{ if} 0 \leqslant x < 1\\
4 - 2x & \mbox{ if} 1 \leqslant x \leqslant 3
\end{array} \right. $

Answer

Absolute minimum: $f(3)=-2$. No local minimum. No absolute or local maximum.

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Video Transcript

we will sketch the graph of the function given by this expression, the piecewise function the first part or piece is two X plus one. For x Greater than or equal to zero and less than one. And the second piece is 4 -2 eggs. If x is greater than or equal to one and less than or equal to three. So it's a piecewise function with two pieces. Both pieces are straight lines. This case they will be segments. And the domain of the function defined that way is zero three. Close interval, close both. And points in the first piece series included but one is not included and in the second piece both one and 3 and include are included. So let's uh draw first to line two explosive one for X greater than or equal to and less than one At zero. The line to Express one has the value one. So and it is included. Yeah. So we have this value here. This red dot here. Okay then at one it is not included but if we calculate the value is one of the piece to express one we get three. That means that at one we will have divided three but we indicate is not in good with this open circle here and then we draw the line Between those two points. Let's see if like you fitted well here almost there. So you're here this is a line it's a segment that because um x is uh in the interval ceo one including Sedona including one and now the second part is second piece straight line again And for X Equal one, the images 4 -2 which is two. Yeah, so to at once or the values too. So this and it is included in the graph. That is these points here. Okay. Say said we are including it in the graph and for X equals three, we got four minus two times three. That is four minus six. That is negative two. So three we have negative negative two and it's included in the gravels. We have to join the two lines and we see that we have to cross the X axis. That too because at X equal to we get zero. So we draw the line something like this then we um yeah feed better. So there there's a line is a little bit longer than we need and that's it. So this is a line and we see it cross exactly at Mexico's two because the image of two following the second piece is 4 -4 is considered. So we have this uh this continuous function here and we can see that we have an absolute minimum value over here at this point. It's the lowest point of the graph and is included in the graph. That's why it is the lowest point. So f has an absolute minimum value of negative two. In that value. Of course at X equal three. Mhm. Respect to local minimum values. We don't have local minimum values because for example, this point here we have now function to the left of that point. Could not be a local minimum At this point. one here It's not a local minimum because if we take any interval around one Small interpret that is, we are close to one. There are all their images that are smaller of the image at one which is to to the right, we have a smaller values and to the left we have greater values in these other parts of the functions. So one is not a local minimum and there are no more possibilities of local minimum. So have has no local minimum and we respect to the maximum values. There is no absolute maximum because the highest point should be this fine but it's not included. So there is always created values that uh get closer to the image at uh three at the value three but never reaching it. So it has no absolute maximum but it don't have also local maximum because the same reason before there is at any point that we choose on the graph inside the domain, not the end points. There are always images that are greater and smaller than the image at the point. So it has no absolute or local maximum values. So the only thing that this function has is an absolute minimum value -2 at a sequel three at the writing point of the domain. Yeah. And so this is the result in this problem. Mhm