Enroll in one of our FREE online STEM summer camps. Space is limited so join now!View Summer Courses

Problem 48

a. Identify the function's local extreme values i…

01:56

Need more help? Fill out this quick form to get professional live tutoring.

Get live tutoring
Problem 47

a. Identify the function's local extreme values in the given domain, and say where they occur.
b. Which of the extreme values, if any, are absolute?
c. Support your findings with a graphing calculator or computer grapher.
$$
h(x)=\frac{x^{3}}{3}-2 x^{2}+4 x, \quad 0 \leq x<\infty
$$

Answer

(a) $\longdiv { \text { no local extreme value } } ( b )$ absolute minimum at $x=0$


Topics


Discussion

You must be signed in to discuss.

Video Transcript

Okay, so for problem 47 we have the function a shacks which is defined by X cubed, divided by three minus two X squared and plus four X Where domain iss from zero to infinity. Okay, now you are a We need to take the derivative of dysfunction. We have X squared minus four acts us four, and we let this to be zero. Then this is it. Equivalent to X minus two squared, which is equal to zero. And our route will be too. And so, h two, we'll be a loco extreme. Now, is this a, um next thing to do is to identify whether this value is is an absolute extreme. Um, the way to do that, just a first. We need to use directive and let the dream, too, to be bigger than zero and find out the the increasing interval. So seems this expression can be transforming to exploit it to where, and this is always an elective on. So that means our increase interval will be from 0 to 2 union to to infinity. We have to ethics have to exclude to in this case, because when X is equal to two, then our expression will be zero and that it is not, um, bigger than zero anymore. So that means our function will be like this. It's from 0 to 2 and two to infinity, so it is always increasing. It's kind of like that should be like this and nothing too. Uh, another thing to notice. Instead, the function here has the local extreme, but it is not. It is not an absolute extreme, because the function is it's increasing from two to infinity and well goes to infinity. So there's no upper bound and sing reason. There is no there were bound to this, uh to dysfunction or the minimum value here. The the minimum value for or dysfunction on this on the interval from zero to infinity is at zero and so two. It's not an absolute extreme. So that is our conclusion up to is not absolute.

Recommended Questions