find-the-local-maximum-and-minimum-values-of-f-using-both-the-first-and-second-derivative-tests-wh-2

Question

Find the local maximum and minimum values of $ f $ using both the First and Second Derivative Tests. Which method do you prefer?

$ f(x) = \frac{x^2}{x - 1} $

Fahad Paryani · Accepted Answer

Okay. So living has to find the local men in Macs of F using both the first and second order tests. So we&#x27;re large to first use it with the first real test, we take different derivative. And since this is a caution rule, you&#x27;LL apply the quotient room. I mean, since this is a question you&#x27;re going to apply the questionable for derivative. And once you do, you get a simplified answer of X square minus two X over X minus one squared and then very bad, too. And then what we&#x27;re going to do, there&#x27;s we&#x27;re gonna should pull out the factor, X, um, pull out a factor X, you know, give you X Times X minus two all over X minus one squared. We set this equal to zero. Uh, since doing the top part of the function matters, we know that we said the topic go to deal. So we get X equals zero and two, and then we&#x27;re going to do is we&#x27;re going to create a sign chart and evaluate how the sign is changing for our function since X minus one squared is always positive. We&#x27;re not going to include that in OK, So then leave and X minus one we&#x27;re gonna do a zero and two, and then we&#x27;re going to do is look at how the signs are changing. So for all members lesson Joe exit negative. And then there&#x27;s the rest of it is positive. And then for all members lessons, you&#x27;re X minus to the wives like Well, why weren&#x27;t one X minus two is negative, And then this negative and the positive shouldn&#x27;t. Now we multiply to get the sign of the prime. So negative times negative is positive. Pilots have negative negative and positive. Positive. Positive? No. Now what we do is, uh we&#x27;re going to see how the function is increasing or decreasing, so we know it is increasing and decreasing, and then it is decreasing in an increasing. So by looking at this, we can ah one. And this one is going from positive to negative. We have a local max, the local Mac at X equals zero, and then local men, when it goes from negative to positive at X equals two. So then, for the second derivative cast, what we do as we evaluate the second derivative at the points in which the first driven in Syria should I mean two seconds? So first, we have to first find the second every letter that comes out to over X minus one cube. Um, and we evaluate at zero and two. So f crime after crime is you. We get to over negative one. Q was you&#x27;re still negative juices and negative, too. Men at half price to get positive, too. So when have promised lesson zero, we got conquered down, which means it&#x27;s a local Max. And then when it is caught him in, when it is positive, his Khan caged up, which means the local men. And that is how you find the local men and Max using the first and second division of the test.

Problem

Find the local maximum and minimum values of $ f …

Problem 20 Medium Difficulty

Find the local maximum and minimum values of $ f $ using both the First and Second Derivative Tests. Which method do you prefer?

$ f(x) = \frac{x^2}{x - 1} $

Answer

$f$ has local minimum at $x=2, f(2)=4, f$ has a local maximum at $x=0$ ,
and $f(0)=0 .$

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Video Transcript

Calculus: Early Transcendentals

Grace H.

Catherine R.

Samuel H.

Joseph L.

Integration Review - Intro

Definite vs Indefinite

$f$ has local minimum at $x=2, f(2)=4, f$ has a local maximum at $x=0$ ,and $f(0)=0 .$

Calculus: Early Transcendentals

Grace H.

Catherine R.

Samuel H.

Joseph L.

Integration Review - Intro

Definite vs Indefinite

Grace H.

Catherine R.

Samuel H.

Joseph L.

$f$ has local minimum at $x=2, f(2)=4, f$ has a local maximum at $x=0$ ,
and $f(0)=0 .$