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# 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). $f(x) = \frac{1}{2}(3x-1)$, $x \leqslant 3$

## Absolute maximum$f(3)=4$. No local maximum. No absolute or localminimum.

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here. We will sketch the graph of the function one half times three x minus one. For x less than or equal to three. And we'll use that graph to find the absolute and local minimum and maximum values of the function do we have f of x equal one half times three X -1. And ex belongs to the interval negative infinity. Up to three Including 3. So we know this is a line for that. We need to draw the line, we only need to point and in particular at The writing point of the interval three the value of the function is F at three is 1/2 times three times 3 -1. He's 1/2 time is 9 -1 is eight over to equal four. So f at three equal 4. That's a result. We are interesting because yeah, um this is the writing point of the interval. So we used that value to draw the line. So we put the value here. This point is the image of three is for here And then we need another point. For example one when x equal one. If at one is one half times three times one minus one. And that is one because inside parenthesis we got three minus one is 2/2 is one. So at one we have image one as we see here and with that we have completely defined the graph of the function and there is uh and increasing behavior of the function But if we go to the left function so will descend in this way. So this is it. This is a graph and photograph we see easily that we have effectively and Low and high. His point on the graph which is the image of three, which is four. So f has an absolute maximum value four which we calculated here in that value of course At x equals three which is the right and point of the interview after the main. Yeah, okay. But that's all there is to it because yeah, there is no local maximum because if we take any point on the graph there's always points or numbers with images larger and smaller than the image of the point. In particular the writing point of the interval, that is this point here, it's not uh a local maximum because we have no graph to the right, even though the function could be different there, we have taking this domain. So we have no local uh maximum at the radio point and at no other point. So there is no no local maximum. Yeah. And in regard to the absolute minimum there is no absolute minimum at all for this function in this domain. Because functional solving increasing that is if we move to the left this direction here, this graph is always the sentence. So there is no lower bound to the graph of the function. So f has no absolute minimum but there is also no local minimum because at any point there are images larger and smaller than the image of the point. so absolutely, or local many months. So this is the the result. The only element we found is an absolute maximums of the function, which value four. And that, of course, at the writing point of the interval X equals three, and that's all for this function.

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