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4. flr) = 4ri-rif(r)f"W)critical values:local mar:local min:inflection point:Determine the intervals when f(x) is:increasing:decreasing:Graph:concave up:concav...

Question

4. flr) = 4ri-rif(r)f"W)critical values:local mar:local min:inflection point:Determine the intervals when f(x) is:increasing:decreasing:Graph:concave up:concave down:

4. flr) = 4ri-ri f(r) f"W) critical values: local mar: local min: inflection point: Determine the intervals when f(x) is: increasing: decreasing: Graph: concave up: concave down:



Answers

Find the intervals on which $f$ is concave up or down, the points of inflection, the critical points, and the local minima and maxima. $$ f(x)=x^{2}(x-4) $$

Mhm. They even problem. We want to um determined where the intervals are concave up or down the point of inflection critical points and the local minimum maximum. So what we're going to do here is we're going to look at our function F of X equals one over two. For and when we do that, we first are going to notice the fact that There's a local maximum right here at 01. But if we didn't use the trace feature, we could also just look at the prime of X&C. That there's a maximum reached when X is equal to zero. So then we just plug that in here and see that I have the vaccine equal one. Then we see that that's our critical point. We also have inflection points. There's gonna be three inflection points. So the graph is going to be concave up. Think on cave down, then concave up. So we see that there's going to be the three inflection points because of that. Mhm.

We are given a function X. The fourth minus four X. We want to find our extreme A. Our regions of can cavity and our inflection points. 1st. The extremist. We should note that the derivative is for x cubed minus four. Our critical numbers will be where that zero or non differential but it's never not differentiable. So that set it equal to zero. Okay, we could add the four to both sides, Divide by the four. And then if we take the cube root, we get the solution For a critical number of x equals one. Check on both sides of that to see the sign of the derivative by filling zero. I get a derivative that's negative. If I fill in two I get a derivative that is positive. So we should see that X equals one is giving us a local minimum. And if we fill that value of one in for X. Into the original function will get negative threat. Next for can cavity let's look at the 2nd derivative. That second derivative is 12 x square. Okay, so possible inflection points would be where that is equal to zero, Which very obviously is when X is zero. Let's check the regions of the cavity around that. Okay, If we fill in a negative, what is it? Well, it is positive, Fill in the number greater than one, it's still positive. So I have concave up, followed by concave up. So this is always concave up. Concave up from negative infinity to infinity. And there are no inflection points because there was never a switch in the cavity.

Okay for the given problem, we want to find the given information or the needed information based on the function F of X Equals X to the 3/2ves -4 X to the -1/2. Mhm. So with this in mind where X is going to be greater than or equal to zero, we want to determine the uh local minimum maximum. First. We see that there is not really going to be a local minimum or maximum. And we can also tell us by looking at the have prime draft, we see that it never reaches zero. So that means that there is no local minimum or maximum that being said, we do see that there is going to be an inflection point because if we take F double private acts, you see if there's going to be a point at which the graph equals zero, that's an inflection point at two. And that will be tell us that our graph is concave down from negative infinity to to or from 0 to 2. And concave up from two to infinity.

For the given problem, we want to find intervals on which fs concave up or down the point of inflection. The critical points and the local minimum maximum. So first we're going to start off with F prime or just F of X equals x squared minus x cube. Or you could use t whichever you prefer. We see that the local minimum and maximum can be determined by the um the initial function. You see, we get 00 and this point right here. Or we can always just take f prime of X and see that these are the X coordinates at which the maximum minimum will occur. And then we just plug them back into our original equation. Then this also gives us the critical points. So now we can take f double prime of X to find the inflection point. We see it's going to occur at one third. So that means that from negative infinity to one third. The graphics concave up and then from infinity or from one third to infinity. The graphics concave down


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