Enroll in one of our FREE online STEM bootcamps. Join today and start acing your classes!View Bootcamps

Problem 31

a. Find the open intervals on which the function …

Problem 30

a. Find the open intervals on which the function is increasing and decreasing.
b. Identify the function's local and absolute extreme values, if any, saying where they occur.
K(t)=15 t^{3}-t^{5}


See the graph



You must be signed in to discuss.

Video Transcript

So you have KFT is fifteen t cute minus t to the fifth. And if we take the derivative to find the critical points, we get forty five t squared times five t to the fourth, and we can actually factor This has five. And then that'LL be nine. Well, we can actually thanked around a t Square, too. Five t squared, And then we should just have Yeah. See, You may have nine t squared him. Sorry. Nine minus t skirt area. And so if you want Kay proud to be zero to found a critical points, we can factor this even further. Sits five t squared times could be three minus t times two b plus t. Yes. Zero instead are critical ones are zero three and maggot is three. All right, let's draws the number line. We have Kludd negative. Three, zero and three and we'LL look a k prime. So it's the left of negative three. Well, this is always going to be positive. So to let the negative three this's going to be positive and this is going to be negative. So even naked negatives, and in between negative three and zero, what happens this is still going to be positive, and this is going to be positive. Should be positive. And then it zero way don't change signs. Were soldiers positive and then positive? Positive. But then, after three, this term becomes negative. All the other to be stay positive. So the negative. So then Kay is decreasing between negative infinity the night of three. Increasing between nature three zero, it's going to be increasing from zero to three and then decreasing again from three. Okay, so we kind of want to look at a function like this and, uh, and still get back down. Yeah. Okay. So we're gonna have no absolutely extreme because this is a decree five polynomial. It's going from infinity to negative infinity. We are going to have local extreme. So we have a local backs at three, and then we can plug in three and two K to get a value of one sixty two and then we have a local men Negative three in the value we get negative. Three is negative. One sixty two. And again, there's no absolutely extreme