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In Exercises $13-34,$ find the transition points, intervals of increase/decrease, concavity, and asymptotic behavior. Then sketch the graph, with this information i...

Question

In Exercises $13-34,$ find the transition points, intervals of increase/decrease, concavity, and asymptotic behavior. Then sketch the graph, with this information indicated$$y=x^{5}+5 x$$

In Exercises $13-34,$ find the transition points, intervals of increase/decrease, concavity, and asymptotic behavior. Then sketch the graph, with this information indicated $$ y=x^{5}+5 x $$



Answers

In Exercises $13-34,$ find the transition points, intervals of increase/decrease, concavity, and asymptotic behavior. Then sketch the graph, with this information indicated
$$
y=x^{5}+5 x
$$

We're asked to sketch the graph of y equals X cubed plus 24 x squared, using the information about increasing decreasing intervals, transition points, cavity and other things. So let's go through this. We have our function excuse plus 24 x squared. It's derivative is 48 x plus three x squared. You could then find the second derivative. He was 48 plus six x, and that gives us a set of transition points. So when um when X equals zero and six minus 16 why prime zero? And when X equals minus eight, why Double prime is zero. So we can take a look at what the values that these points are. So we get why it zero is zero. Why it minus 16 is 2048. Why minus a is 1024 and then asking tactically, we know that because this is an odd function or not power. Also, when I function, we get that, um that the leading coefficient is an eye function. We get that the function Assam totes the infinity as ex course infinity minus infinity has exposed minus infinity, so we can use all that information to make our plot, which I've done here. And so we do is we take off where we have transition points and then we know their values. And then we know we can see if these things are positive or negative in the different regions. So when this region ex prime or why prime is positive by double prime is negative here both. Why Double primer y primer negative here? Actually, that one is messed up here in this region. Ex prime is negative. Why? Prime Double primer is positive. And then in the final last region here, they're both positive. So we can we can go through and we can sketch the curve that follows between these points with the slopes and the con cavity as real determined, and so I could look something like that.

In this case we want to plot Y equals x cubed minus three X plus five. And so we can right that we can take the derivative get three X squared minus three, take the second derivative gets six X. And we can figure out their transition points when um when y primary zero, that happens at plus or minus one. And when y double prime pre zero happens when X is zero. So from those we can get our Why values at each transition point. So at -1, we're at seven at one, we're at three and zero, we're at five. And then again, the leading coefficient here has an odd power or the highest highest order coefficient is an odd power. So that means that as what expose infinity, why goes infinity and X goes to minus infinity, why goes to minus infinity? We can then make our plot again, we have our transition points here -1, 1 and zero in this region. Why prime is positive? Y, double prime is negative. Here they're both negative here, why prime is negative, Why double prime is positive and here they're both positive. And so we know that here we're at seven, here we're at five and here we're at At three, you can notice that this starts at three and not zero. So we can draw a curve. We know it's concave down in this region and then there's a transition here an inflection point and now it's concave up and we go off to infinity

In this case, we want a lot. Why equals X cubed minus three X plus five. And so we can right that we could take the derivative get three x squared minus three Take the second derivative gets six X and we can figure out their transition points. When um When? Why Prime a zero that happens at plus or minus one. And when? Why? Double primer zero happens when x is zero. So from those we can get our why values at each transition points what minus one were at seven at one worth three and 05 And then again the leading coefficient here has an odd power for the highest highest order Coefficient has is an odd power. So that means that as what ex course infinity Why does infinity and X goes to minus affinity? Why goes to minus infinity? We can then make our plot again. We have our transition points here minus 11 and zero in this region. Why? Prime was positive Why Double prime is negative here. They're both negative here. Why? Prime is negative. Why Double prima's positive here They're both positive. And so we know that if you were at seven Pure five and you were at at three. Can notice that this starts at three and not zero, so we can draw a curve. We know what con cape down in this region. And then there's a transition here. Reflection point, and now it's kind of gave up. We go off to infinity.

Now I want to plot Y equals x squared minus four X cubed. We can getting the first derivative. Yeah. And then the second derivative and set these two things equal to zero to get our transition points Are transition points are zero and 1/6. That's when y prime is zero and when y double prime is zero is at 1 12. We can then plug those values in to get our are why at those points. So at zero at X equals zero, Y is zero. When X is 1/6 Y is 1/1 08. And why is 1/12 X? When x is 1/12 Y is one over 2 16. And again we have an odd meeting coefficient so that we get but there's a negative sign in front of it, so positive infinity. Why it goes to minus infinity and when it goes to negative infinity why goes to positive infinity? So we can plot our curve Again, we have our three for the transition points. And then when this region when x is negative, we have the first derivative is negative, second derivative is positive. So a descending curve with positive with concave up next region, they're both positive. So it continues up like this and the next region the slope is positive but the um the curve is concave down and then the slope changes and his curve curve is concave down. So we wind up with this something that looks like this that goes off to infinity as X increases to plus or minus and


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