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Point) Let f(x) = -7 maxima (minima)x' . Find - the Open Intenvals whichIncreasing (decreasing): Then deterina tha x-cooidinatesUelelsIncreasing Interval $ dec...

Question

Point) Let f(x) = -7 maxima (minima)x' . Find - the Open Intenvals whichIncreasing (decreasing): Then deterina tha x-cooidinatesUelelsIncreasing Interval $ decreasing on tha (ntervals The relative maalmu occur at X = The rolative minima 0t occuNotes: In the first twO, your answer should e ther be "nona" single Interval SuCh(0,1) comma soparaled Iist of Intervals, such (int; 21, 03,41 or tha wodIn the Iast two, your answer should be comma separated Ilst 0t NaluasLne Wor nonee

point) Let f(x) = -7 maxima (minima) x' . Find - the Open Intenvals which Increasing (decreasing): Then deterina tha x-cooidinates Uelels Increasing Interval $ decreasing on tha (ntervals The relative maalmu occur at X = The rolative minima 0t occu Notes: In the first twO, your answer should e ther be "nona" single Interval SuCh (0,1) comma soparaled Iist of Intervals, such (int; 21, 03,41 or tha wod In the Iast two, your answer should be comma separated Ilst 0t Naluas Lne Wor nonee



Answers

For each function, find (a) the critical numbers; (b) the open intervals where the function is increasing; and (c) the open intervals where it is decreasing.
$$f(x)=x^{6 / 7}-x^{13 / 7}$$

So, as you can see, I already have out off schedule. Figure 4.39 which is given in textbooks just like this. So this corner born is at X equal to 10. An hour in points are zero and 20 and value off zero is within. So you want to look at what happens when X is between four and 12. So let me was black color for the first part. So when X is between four and 12 roughly, we are in this situation because we can see we're looking at this part off the function. So So the maximum occurs at left and born and minimal. Is that a critical point? This is a critical point, because derivative at this point is more different. So this statement would correspond to sickened state. This is a statement. I'm sorry. This a statement for because maximum is that left and warned not the right and burnt Maximus that left. So there's a statement for Okay, let's look at what happens when exes between 11 and 16 and let meetings the color to red 11 to 16. They're dropping off this portion. Okay, as we can see. Well, this is quite straightforward. So many Margo's that left and wound and maximum occurs at right and point. So this is your statement. Number one, Let me change the color for part three. So we're between four and nine. So this part four and slightly above the critical point. So for a nine again, this is very similar to part two. So here your maximum occurs at left in point and many markers. Right? And born. So this is straight my number three. Let me James Sekulov apart for And? And we're between eight and 18. You're dropping off this part. We have something like this. So their maximum occurs air, right? And point. And many markers ago. Critical point. So this is your statement. So that's your answer. Thank you.

You think you want to know this about this function is that X can never equal negative to where I needed to is not even in your domain, because we cannot define my zero. Now let's go ahead and find the derivative using B um quotient rule bottom times the derivative of the top minus the top times the derivative of the bottom allover the bottom increased power by one. Cleaning this up should end up with see extras, too. It's gonna be negative. Five all over, exposed to squared. Hey, all right, we're ready to find critical points or party. What you need to do is to set the derivative equal to zero to suffer a But however, there's no solution to this. The derivative will never equal the road because I can't play the negative two in. So this is just never going to happen. There's no critical point, but we can still apply the test for increasing and decreasing at negative, too, because something weird is happening here. So we need to definitely test that point. So for stuffing probably should, right? No critical points on party. Yes. If you are still confused, why think about the equation negative five equals zero. Is there a solution to negative Five equals zero? There's not, which means there's no critical point. So let's go ahead and test what's going on at negative, too. So we're going to go ahead and plug in some values, like negative three to see what's going on there. Click Negative three in going to have a positive over negative. Meaning it's negative. We're decreasing. If I plug in zero or the number to the right of negative too, get a positive mover. Positive. Which means this number will be positive worth above the X axis. Wait, I was plugging it into the role of being plug it into the derivative right here. So let me double check. Negative three. You're free. It's gonna be It's gonna be negative. Plug in zero. That's gonna be naked. All right. Perfect. So this function is going to be decree never increasing, and it's always decreasing. But we can't include the number negative too. So you could technically right the following. If you need to see a graph of dysfunction, you're always welcome to graph it to figure out what the heck is happening at negative to you don't remember in your rational functions

Being heart. I we have 10 at X equals two increases after and decreases after. So we'll circle right with mass. And well, it's a men we know The interval that office increasing on it is gonna be Yeah, grated into decreasing will be lesson to you. So since you guys from countries increasing, this is a minimal. And in part two, we have two zeros, one ads zero and one at one as increasing outside of these You cruising out on in between. That makes zero on Max and one a minimum. Then in three, it is yours at negative 10 and one Baxter black in 10 one and F crime is negative in thes two sections. Positive. It needs to means that you have a max that's native ones that go from increasing, decreasing and at one and then we have men at zero and lastly in part for looks like make it 51 and no sorry. Negative three one in five are zeros green moving either on. So it's Hawes. It is betweennegative three and five, both both intervals and then native in the others. So that means we have a acts of five a man at natives through, and neither as one

Were given a graph of a function F This is the graph for exercise one of this section 4.3. In part, they were asked to find the open intervals on which F is increasing. Looking at the graph, we see that f is increasing on the open intervals one, 23 and from four 26 in part B were asked to find the open intervals on which f is decreasing. So this is going to be that f is decreasing on well from three, 24 and as well as zero toe one in part C were asked to find the open intervals on which f is con cave upward. We have the f is con cave upward on the open interval. Looking at her graph, we see that this is from zero up to to in Part D rest find the open intervals on which f is con cave downward. We see the fs con cave down on the open interval 24 and also on the open interval from 4 to 6. There's a cuss before and in Part E, whereas to find the coordinates of the points of inflection. So point of inflection this is going to be where the graph changes from Con Cave to convex. When we see just looking at the graph, that graph changes at the point 23 so we have that point of inflection is 23 mhm.


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