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Use a computer algebra system to graph $ f $ and to find $ f' $ and $ f" $. Use graphs of these derivatives to estimate the intervals of increase and decrease, extreme values, intervals of concavity, and inflection points of $ f $.
$ f(x) = \dfrac{x^{2/3}}{1 + x + x^4} $
Calculus 1 / AB
Calculus 2 / BC
Chapter 4
Applications of Differentiation
Section 6
Graphing with Calculus and Calculators
Derivatives
Differentiation
Volume
Oregon State University
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we want to use some kind of cast software to graft. The function of X is equal to X to 2/3 over extra four plus X plus one. And then we also want to use this to find the first and second derivatives as well as to grab them. And once we grab the derivatives, we want to bye intervals of the functions increasing, decreasing con cavity as well as any extreme points we may have as well as any inflection points. So I went ahead and graft ffx already. So now I just wanna have a zoom kind of far out and look to see if there was any behavior towards ends on my care about In this case, I think the best window for us to be would just be from negative three, 23 And we get this nice looking curve here, so just kind of looking at this. When we take the first derivative, we know we should have some kind of maximum, maybe around like negative 0.8. We should have a minimum at X is equal to zero. Or it may be undefined at that point because it kind of looks like it's really pointed at that point right there and then we're going to have a maximum around 0.5. It looks like so, no, that we kind of have an idea of where we might have some of our maximum men's. Let's go ahead and see what we get for the first derivative. So I went ahead and plug that equation into a derivative calculator, and it gave us this for our driven it. Now one of the things we're gonna have to remember is so normally for critical values, we just look for where the function is set equal to zero. But we also have to remember that it's undefined. We need to check those points as well, and in this case we have a vertical ask. Himto at X is equal to zero, and so we know that X is equal. Zero could be a possible critical point for us, so let's just kind of look to this because we go back to our original one. We see that this is going to be defined for all real numbers, so let's go ahead and find where the function is increasing and decreasing now to the left of negative 0.722 The function should be increasing because remember, Yeah, Prime of X is greater than zero. That tells us we're on an interval. Where is increasing? So to the left, it will be increasing. And to the right of that point, it'll be decreasing. Which tells us we should have a local Max at X is equal to negative 0.722 about now for other point over here, what with increasing into the point and then decreasing after. So again, we're gonna have a baby backs here and now about zero. So I'm just gonna go ahead and write this over here about X is equal to zero. So to the left of our horizontal asked are vertical ass into the function is negative. So it's decreasing into ecstasies of zero. And after it it is positive. So here that we get at X equals zero. We have a local minimum. Now that we know one of values for X or maximum mittens, let's go ahead and write down these intervals so f of X is greater than zero. So it looks like starting from negative infinity all the way up to our first exercise. So negative zero points of you, too. And then it becomes posit again after zero, and then they'll stay positive until about 0.611 So that's our interview over the functions increasing, defined, decreasing. Remember, that is one F prime of excess Tripoli less than zero. And that's just going to be the rest of our interval here. So from negative 0.722 up to zero and then union 0.6112 and Kennedy Right now, let's go ahead and check out our second driven it and try to find some con cavity. So just like our first revenant, we're going to have this vertical ass in tow at 0 to 0 now. So we may need to check to see if this is an inflection plans. Well, let's first check these four X intercepts that we have. So to the left of negative 0.966 it looks like our second derivative is positive. So remember F double prime of X, strictly greater than zero means we're going to be Khan caged up. So where up. And then after that point, it's down. So that means here we will have a point of inflection because con cavity changes our next point. Woods con cave down to the left and gone gave up to the ring. So point of inflection, our next zero, which is right here. Let me go ahead and do this one up here, though, it's Kong cave up to left and concrete, down to the right. So that's also a point of inflection. And then over here at 1.115 it is Kong Cave, not comfortable but concave down because it's negative to left of it. And then Kong cave up to the right, so that would also be a point of and deflection. Now I'm just like before. Let's go ahead and check out zero so X is equal to zero. So to the left of X is equal to zero. The function is negative and also to the right of X is equal to zero. The function is also negative, so that means it's concave down on either side. So that means this here is not a point of inflection. All right, so now let's go ahead and actually write hour intervals. So this is gonna be negative infinity to our first X intercept 0.966 And then we'll union that with negative 0.4562 negative 0.123 And in our last interval of where it is positive will be one point 115 to infinity, then went f the will find a strictly less than zero. We know the function would be concave down and this will occur on negative 0.966 to negative 0.456 union with negative 0.123 and the least amount a little bit 20 and then 0 to 1.115 So we have our intervals where the functions can keep up, Con caved out and all of our points of inflection So we're done with the problem.
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