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Ma N.

November 6, 2020

f(x) = 3x 2 ? x + 2,

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### Video Transcript

Okay, So to solve this problem we need to do is take a look at the graph that's provided in the book and find out what the values are at certain points. So, in part, A says state the value of that one. And we look at the graph here, it becomes apparent that when X equals one, I never win as three. So everyone as a value three now half of negative one. We just looking it over here. And if you look at your book, which provides a little bit better resolution, I say it's about thinking of zero point two five now. Parsi asked for what Values of X is of of X, equal to one. And we see that f of X is equal to one when X is equal to zero here and also when x is equal to three right here. So for one values of X, that would be one. And also I'm sorry zero and also three. Let me just write that over again. Carol. And so, as they made the value of X, such that F of X is equal to zero. So when F of X is equal to zero That's when it crosses this point right here, friend, when it crosses the X axis and that looks to be about perhaps around zero point seven five again, you should refer to your book since it's a little bit clearer than this drawing here. So the domain in range So domain is basically what values of excess car covert are defined. Can we see here that domain is from negative to two four, and we can use the brackets here too, to show that the points are indeed to find out negative too. And at four in the range of this, when be from again using the scribe records Negative. One, two, three. So we noticed that it Starks. The lowest value of the function is that negative one and the highest one is at three. So that is the domain and range of the function. And the last sub question here part have passed on what interval is increasing and we could see that increases from negative too. Two, one grandson from negative to tow one, we can see that the graph is indeed increasing. It has a positive slope. So the answer to part ofthe ISS from negative two two one and that should cover pretty much the question here

University of California - Los Angeles

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