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University of North Texas



Problem 16 Medium Difficulty

If $ f $ is the function of Exercise 14, find $ f' $ and $ f" $ and use their graphs to estimate the intervals of increase and decrease and concavity of $ f $.


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Video Transcript

So in exercise 14 we worked with function F of X is equal to two x plus three squared times next, minus two to the fifth, All over X cube time X minus five squared. And what we want to do is to find the first and second derivatives, preferably using some kind of cast software and then use their grafts to estimate intervals where the function is increasing, increasing and buying contact. Body of All right. So, um, when we actually worked out exercise 14 you just using the graph of Ethel X two. Get approximations. We know that we should have minimums at X is equal to negative three. House and X is equal to about eight. So we're just gonna keep that in my just to make sure we get the same values for this. No, we should have a vertical ass into it right here at X is equal to zero as well. Over here, X is equal to five. So she's going to put those in. And I just went ahead and already found a crime using a derivative calculator because doing that by hand would be way too much work for any sane person. And then I went ahead and graft it, and I made sure I included all my ex intercepts and all of that. So with this here, you will see that. So this here was around Negative three hats. This here is that, too. And then over here, that value was about eight. Now, to figure out where the function is increasing, we want to determine that F prime of X is strictly larger than zero. So this here will go off to date, if any. If you were to look at the rest of the graph to the left of negative three house so to the ray is gonna be negative. 3/2 to our first vertical awesome being zero in union and then from zero to four or too excessive. 505 are next intercept. It will be increasing. And then after this, we will deposit from eight to and and maybe I should just roll that over here. The behavior of it looks like that. So it's decreasing or it's negative to the left of eight and then pause it to the right. All right, so we have our interval of increasing. And now to find where it's decreasing and let me go ahead and write that about here. This is increasing now where the function is decreasing. That'll be where prime of X is strictly less than zero, and that would just be the rest of our interval. So from negative 22 negative three hops union and then from 5 to 8, it'll be decreasing. And so let's just go ahead and check that the derivative here actually does give us two minimums at exited with a negative three house and about eight so into negative three house. The function should be decreasing and then increasing. Doctor. So that is a minimum at X is equal to well, it's increasing into the point and then increasing after. So that's gonna be a saddle. So we don't have any issues there, and then at X is equal to about eight. Well, it's decreasing in the point and an increasing after. So at least our graph here matches up with what we would expect for the maxes and meds. All right, then we want to find Kong Cavity. So began for the second derivative. I just went ahead and plugged it into a derivative calculator, and it gave me that. Ah, big, monstrous looking thing. And then I just kind of zoomed out far enough to kind of get a good idea of what our behavior should look like. And this is what I ended up with for my graph. So for us to find Kong cavity, remember, we want to figure out where the function he's going to be larger than zero for it to be Kong cape up. So f double time of strictly larger than zero. Uh, and again, I went ahead and put in the vertical Assam coats. Just we kind of keep track of them. So it looks like from 0 to 5, a double prime not to bury. You're five. It looks like it is positive. And then on the other side of our asking toe, it will be positive. That's what. So it's concave up from 0 to 55 30 then forward to be Kong cave down. We want to find where the second derivative is less than zero, and that would just end up being from negative infinity to zero. And if you were to go ahead and actually use this information or at least used to look at the work we get in number 14 you'll see that it pretty much matches up pretty well with what we have here.

University of North Texas
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