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Use the guidelines of this section to sketch the curve.$$y=x \sqrt{2-x^{2}}$$

$(0,0)$

Calculus 1 / AB

Chapter 4

APPLICATIONS OF DIFFERENTIATION

Section 4

Curve Sketching

Derivatives

Differentiation

Applications of the Derivative

Harvey Mudd College

Baylor University

University of Nottingham

Lectures

03:09

In mathematics, precalculu…

31:55

In mathematics, a function…

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Use the guidelines of this…

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we want to sketch the curve. Why is it into X times this word of two minus x squared. So in the structure they give us this laundry list of steps we should follow any time we want to sketch something, and so let's just go ahead and start with that. So the first thing we want to do is to determine our domain of our function. And so that's going to just be where, under our radical, it is strictly greater than or equal to zero. So not strictly bury them, but our eagle to greater than or equal to zero. So if we go ahead and factor the inside there, it might be easier to see what that should be. So is going to be the square root of two minus sechs times the square root of two plus X, and so from that we can conclude our domain. You should be the negative square root of two to the square root of two, including each of those points since we can, how under the radical equal to zero. The next thing we want to find is where our function is going to equal zero or our intercepts I should say so let's find the ex intercepts first, So X intercepts. So this is gonna be where zero is equal to X times two minus x squared. So looking at that factor form over there, that would be well X is equal to zero or X is equal to whatever makes underneath the radical equal to zero, which would be ex physical to plus or minus the square root of. And since X is equal to zero here, we know this is our why intercept as well at the origin. The next thing we want to look for is symmetry, so stuff like this isn't really known for being periodic, but we can at least check to see if it's going to be even or odd. So how the negative X is equal to negative X times. The square root of two minus will negative X squared is just going to be expired. So it's when the square and this here is equal to negative effort back. So we know our function is odd, and thus it has symmetry about the origin. So if the draw what's going on from 0 to 0 to square root of two we could just go ahead and reflect that across the origin to get the rest of the ground. What? Now that we have that we can go ahead and look att passin toots of this s troops. But there are going to be no acid toes because we're going to have no horizontal fascinated. Since we do not go to infinity and they and we have no vertical acidosis, there's nowhere in dysfunction that we're dividing by zero. So the next thing we want to look for so steps five and six is where the function increases and decreases, as well as any local maxes and men's we may get. So for that, we need to figure out what wide prime is going. So let's go ahead, get down another page. We're not squeezing things in, So why is he x times the square root to minus X work? So take the derivative This we're going to the team product will remember. Product rule says you take the 1st 1 will comply by the derivative of the second, and I'm going to rewrite that as a powder so two minus X squared to the one, huh? And then we need to add it in the opposite order. So to Maya's X squared times, the derivative of X. So take the derivative of two x minus minus expert to the one happen, going to need to use power and shingle. So it's going to be one house to minus X squared to the negative 1/2 times the inside, which is going to be negative to X and the derivative of X rays. Just want to be one now for usable but algebra weekend. Rewrite this as two times one minus x weird over the square root oh, to minus X squared. Now we set this here equal to zero. That's going to tell us that one minus X. Where is Eagle? Deserve or X is equal to plus or minus one, So these are possible critical points and Millis. We also get critical points when our denominator is equal to zero or X is equal to plus reminds this word, but since these are endpoints, we don't actually need to be concerned with them. But if for some reason these weren't in points and the function was defined past this, then we would also need to look at those as well for critical values. All right, so let's go ahead and write that information down. So we have two times one minus x word over to minus X squared square root. Now we need to figure out where this function is increasing and increasing. So that's going to be where. Why Prime is strictly larger than zero, and this means the function is increasing. And I already went ahead and solved this beforehand. This being the interval negative 1 to 1, and to figure out where the function is decreasing, we want to see where wide prime is strictly less than and again already did this beforehand. And it's just going to be the rest of our domain that we have here. Excluding are in points so negative square root of two to negative one union, one to the square root. Now let's go ahead and look at that critical point that we had, which was are you actually have two critical points ex busy with the negative one and X is equal to one. So to the left of X equals negative one. The function is decreasing and after one, the function is decreasing and between negative one and one, The function is increasing. So what this tells us is X is equal to negative. One should be a local men and exiting Goto one should be eight local Max. All right, The last step before we actually start grabbing bits is to find our con cavity and any possible points of inflection. So we need to figure out what why double Prime is going to be. So let's go back over here now to take the derivative of this function we're going to need to use. Quotable. So why don't fine is going to equal to. So I'm gonna first factor that two out front so I don't have to carry it around and question Will says Low D hi and the numerator. We had one minus squared, I thought red there and then minus hi below. So them in the opposite order Thanks Oh, to minus X squared rooted all over our denominator squared. So that would be to minus X squared. And we would absolute value this but honor domain that will always be stricken, Arjuna zero. So we don't need to worry about that being zero. So now the derivative of one minus X squared what we said at the last part that that should be negative. Do X And we also found what the derivative of square root of to my sex squared is in the last part which would just be 1/2 to minus X squared to the negative 1/2 times negative to x and once again using a little bit of algebra. To simplify this down, we will get that this is too x expired minus three all over, two minus x squared to the three house help like And if we work, is that this equal to zero? We're going to get that X is equal to zero or X is equal to or zero is equal to X squared minus three, which would imply exit what plus or minus the square root of. But this here. So the square root of three is strictly larger than the square root of two. And negatives were 23 strictly less than the square root of too negative. So that tells us these are undefined points and we don't need to worry about them. So the only actual inflection point we get is exiting zero. And again we could set her denominator equal to zero. But we would just get that y double prime would be undefined at X is equal to plus or minus square root of two. And just like over here in the first part where we found our critical points for Why prime, since they're in points, we wouldn't really care about them being critical, since we can't really say what happened on the other side of the function. So only point of this in court. Inside of the second grave, it will be X is equal to zero, right? So, again we got our second derivative was two X over X squared minus three all over to minus X squared to the three house power and not a little bit. Now we need to figure out where the function is concave up, so they'll be calm. Cave up when y double private strictly larger than zero. And I would have been solved this already, and this should be on negative square root of 2 to 0, at least over our domain. It should be that. And it's con cave down when y double prime district, less than zero or zero to the square root of now the only possible point infection that we had was X equals zero, and we can see to the left. The function is concave, but and to the right, the function is concave down. So since we have this change of kong cavity, we will know that that is a point of inflection. All right, so that was everything they told us we should do before we actually start sketching the curtain. So let's go ahead and start putting down our intercepts. So we haven't intercept at zero, and we haven't intercept at the square root of two and negative square root of two. We know the function is going to be symmetric about the origin, so we only really need to draw it from zero to square two. And then we can just sketch the curb in the opposite manner going, um, to the negative square. We have no accent. Oops. We know we have a max at exit one. So here is a max, and on the other side, negative one should be Amen. And we know what X is equal to zero. We should have a point of inflection. All right, so let's go ahead and start it ecstasy. Cool zero and go to the square root. So it is increasing until we hit one, because that's our maximum. So it's going to come over here and then fly and out at Exeter one. And then the next important point is it starts decreasing and we hit our intercept here and now we will just go ahead and reflect this across the origin, insuring that negative one. We have our minimum and be hit our intercept. So this year would be a sketch of our graph. And just to make sure so at excessive zero we do have are changing cavity because to the left is calm cable and to the right, it's Kong cave down. Now the only other thing you might want to possibly go do is to say, with this actual minimum value and maximum values should be. But other than that, since we're just sketching get, I think this is fine, since we show that it is a maximum at one as well as a minimum at exit

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