Like
Report
Numerade Educator
Problem 17 Medium Difficulty
(a) Find the intervals on which $ f $ is increasing or decreasing.
(b) Find the local maximum and minimum values of $ f $.
(c) Find the intervals of concavity and the inflection points.
$ f(x) = x^2 - x - \ln x $
Answer
(a) $f(x)=x^{2}-x-\ln x \Rightarrow f^{\prime}(x)=2 x-1-\frac{1}{x}=\frac{2 x^{2}-x-1}{x}=\frac{(2 x+1)(x-1)}{x}$. Thus, $f^{\prime}(x)>0$ if $x>1$
[note that $x>0]$ and $f^{\prime}(x)<0$ if $0<x<1 .$ So $f$ is increasing on $(1, \infty)$ and $f$ is decreasing on (0,1)
(b) $f$ changes from decreasing to increasing at $x=1$. Thus, $f(1)=0$ is a local minimum value.
(c) $f^{\prime \prime}(x)=2+1 / x^{2}>0$ for all $x,$ so $f$ is concave upward on $(0, \infty)$. There is no inflection point.
Discussion
You must be signed in to discuss.
Top Calculus 2 / BC Educators
Catherine R.
Missouri State University
Samuel H.
University of Nottingham
Michael J.
Idaho State University
Joseph L.
Boston College
Watch More Solved Questions in Chapter 4
Video Transcript
Okay, so we're being I have to find intervals in which efforts increasing decreasing local Minden max and into a world of con cavity for after sex. So in order to for find where the function is increasing or decreasing, we take, we apply the first derivative test. We have to take the derivative. So this will come out to be too X minus one minus one over X. And then what we're going to do is we're going to factor out there. One of Rex this cake factory wanted it back. So that means just to act when I'll become to x squared so that it remained to X and then this is going to become minus. Thanks. Come, come minus one. And then now we're gas. A factor to the inside terms and this will come out to be one over x. This factors into two experts corn and X minus one. You said the secret is zero. And at a convenience, this is what is easier to solve Brazil. So we know that one of our exit gone. He was now appointing steady call zero and two x plus one because your X equals negative one half and X minus one that one. However, since our domain of this function due to Alan of X is all ex greater than do, we don't include negative one. So now all we have to do is create a sign charts to see where our function is increasing or decreasing. Since one of her exes, always positive, could run our domains of X rated endure. We can ignore that case So we'LL bring up two x Plus one and X minus one, and we're going to do it from zero and one. We've been over everything lessons really, because I don't mean against Korea on zero when the line across eso for two X plus one, it's all positive and then for X minus one, a number between number between June one gives a negative on the scan. Positive. You multiply these together we get the sign of a crime because there's a negative because that's positive. Now we know where it is. Increasing, decreasing each time. So it is increasing on zero two. I mean, I'm sorry. One to infinity and then we noticed decreasing from between your own one. And since we know ah, the way it is increasing and decreasing. We know the local math and men since it is decreasing and then increasing. We know that there's a local men so local men occurs at X equals one and there is no home Max. And for con cavity we have to take the second derivative second door over there is on two plus one of wrecks where should be too plus one over X squared and we do the same thing without the secret zero. However, as you see that this can never equal zero. So we have to look at whether this is positive or negative. So this is never the case. If you look at this since we are, domain is ex greater than you. We know that F crime has to be on the positive. So since after some kind of positive, we know that it is. Khan came up everywhere honest domains of concrete on zero to infinity, which is a stunning And since there's no change in signed er is no inflection point, no, in a collection point and that is all
Top Calculus 2 / BC Educators
Catherine R.
Missouri State University
Samuel H.
University of Nottingham
Michael J.
Idaho State University
Joseph L.
Boston College
