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Problem 48 Hard Difficulty

(a) Graph the function $ f(x) = e^x + \ln | x - 4 | $ for $ 0 \le x \le 5 $. Do you think the graph is an accurate representation of $ f $?

(b) How would you get a graph that represents $ f $ better?


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Video Transcript

This is Problem forty eight of Stuart Calculus Expedition Section two point two Party graft. The function after effects is equal to either the X plus Ellen time to the absolute value of the quantity. X minus for four. The X values from zero to five inclusive. Do you think the graph is an accurate representation of us? So we take this function amusing, whichever grafting till you may have maybe a graphing calculator or a non linger graphing calculator, and we put dysfunction for this range of zero to five. Here's one example where we see this smooth function, this smooth curve with this dip X equals four. And there's another example. This is another graphing utility that provides a very similar curve. What that small detail at four. Okay, so I may this be, ah, inaccurate and an accurate representation of F. Well, we have to look at the domain of the function, and what we should recall is that the log a rhythm, the natural algorithm cannot take any negative values, and it can also not take any zero input. The absolute value function here around the X minus four ensures that we will not have any negative. Ah, inputs or negative numbers that a lot of natural algorithm has to deal with. S O That is fine. That allows for almost any acts to be plugged in. However, in this range, we see that X equals four provide an issue because X equals two four minus four gives us a zero absolute value of zero is zero. And then we have the natural log rhythm of zero, which is undefined. So we will, right over here. Natural Aga zero is undefined. So this is the issue that is not accurately represented in either the functions right there of the grass in this first graph here, it seems as if there is a point to find here at four. Um seems to be about twenty over here. Very similar results. It doesn't show a hole right at shows a value at for a defined value. And both of those air inaccurate. What behavior we actually should see at four has to do with an infinite lim, actually, because the closer we get to four, the closer that this entire term get closer to negative infinity and quickly we're going to recall the natural order them shape looks like this naturally X. And as you see as we get closer and closer to zero on this x axis, we get closer and closer than negative infinity. And we see that in this function with this absolute value sign the closest and the smallest that this number can being zero and therefore this log rhythm of zero approaches negative infinity. So the point of this discussion is that this function at X equals four should show an Assem tote where the function from the left and the function from the right approach negative infinity and one way we could achieve this one way we can achieve one more paragraph would be too would be to split the function into two parts. There will be FX and there will be a peaceful dysfunction where the function may look like this or minus X when X is between zero and for non including for and very similar to the original function even X plus Eleanor Ellen off expense for from for not inclusive two five. So here's an example of how we can represent exact same function, but in a more accurate sense, because we have excluded X equals for from the defined values because we already discussed that for is not defined. And what we ensure is that the value within the log rhythm is always positive. Hear from zero before we see that we are always subtracting. Remember less than before making sure that this numbers always had to find a value Wait also have the same function here from forty five. This is always a positive number and most importantly, as we approach for from the left, disfunction goes to negative infinity. And as we approach for from the right, dysfunction also approaches negative infinity. And so we would have confirmed and Assam towed here and the function would be more accurately represented in this case.