Problem

(a) Find the intervals on which $ f $ is increasi…

08:38

Problem 12 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) = \frac{x}{x^2 + 1} $

Answer

See Solution

Related Courses

Calculus 1 / AB

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 4

Applications of Differentiation

Section 3

How Derivatives Affect the Shape of a Graph

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Video Transcript

we're being asked to find interim whichever's increasing indications of local Max Amann and Intervals account Cabinet's working one of FX. So in order to first find where effort increasing or decreasing, we have to take the first derivative cat. And so we have to find the derivative of the function. So So this is a question conscience. We apply the question rule. Um, I have already taken distributor previously, so I have the expression ready. So one minus X squared over X square plus one squared and then we're going to do is set the function before zero. So since one minus X first only thing that can contribute to its allowing it to be zero we let one minus X squared equals zero, which then gives us the value of X is equal to plus or minus one. So then we then apply a sign chart and with the horizontal axis representing another line, and the desire is representing our punch of our function. So in this case, only part we're really interested in is one minus x word. Because if you look at the bottom, it is X squared plus y squared, so any negative value will always be positive. So we don't have to really put this and our sign chart. We just have to really look at whether one minus X squared is changing. Klein. So we're looking at negative one and one. No, I don't think that Negative One in one. And so when it is less than negative one, we'LL get assigned us negative. I mean, between negative one and one get positive because their cloaking Joe, you get one and anything greater than one. You get negative and so thiss it essentially tells us the sign of a prime. And since this is Theseus eines of the crime, we know that if we know the sign of a crime, we know where it is. Increasing, decreasing. So when have friends eso where it is increasing occurs when f crime is positive. So there's a curse between negative one and one and that is it. And where it is decreasing occurs between negative infinity and negative one and one and positive infinity. Yeah, And then since we know where the function is ah increasing and decreasing, we can have to figure out local maxim meant So there's increased intention is decreasing from negative finish your negative one and then increasing. We know that increasing decreasing gives us a local men. So a local men occurs at X equals negative one. So ek equals negative one. And then that local max occurs when they go from posits a magnifico. Does it go from increasing to decreasing giving us, eh? A negative? I mean, giving us a little back this Ryker that except for me. So then, for the interval of con cavity and inflection point, you have to take the second derivative. So you have to apply. Since this is a question again, if you apply the quotient rule to this f crime. And I have already taken this to every for time's sake. The second derivative simplifies down too. Two X cube my net six six Oh, minus six x over X cube plus one cube. Okay. And we set. This function is called a zero to find where it is, where the interval them kind of con cavities are. They're doing the top part of the functions contributed. As we said, the top equal to the O to excuse minus six sec. The car is here. We can template fi this by plug by taking out a to X. This leaves us with X squared minus three. And he said this sickle is here and we had excessive Orazio and plus and minus for three. Ah. Then you create a signed shirt again, but this time for planning a second derivative. Now we have our ears. Yeah, negative. With three zero and three and then we are interested and only the top any the numerator of this function. Because if you look at the denominator, um, the X values are being squared first, so any negative number will be positive in any number as positive. Just taking this cube will give us a positive numbers. So we're gonna soon we can just ignore this part of the function. So we're going to look at the top part, and since we're breaking down the top are it could be making a lot you did to evaluate the sign. So we break it down to X and X square minus street, and India's two together will tell us a sign of double prime. So anything less than negative with three gives us a negative number for two X when plugged into to it, any number between a route three injured gives us negative and in European. So this gives us two positive positive and then for X square minus tree any numbers less than X ray remind any number flagged in just less than negative for three into X. Remind story gives us a positive number and then you repeat so between numbers, negative Williams oh gives negative, negative and positive. And then what you do now with multiply across so positive time to negative is a negative negative times. Negative is positive, negative, positive, negative and positive. Positive, positive. So now that we know the signs of the double prime, we also know the interval of con cavity. So when it is Khan, just different color. So when it is con cave up, this's not when it isthe con cave. Thank you. So it looks like this. This occurs when f prime iss positive. So this is between negative on three two zero and with three to infinity and then Kong cave down, which looks like an upside down. You occurs at where have prime is negative. So this occurs our negative infinity to nailing her three and zoot between zero Andrew three and our inflection point The cure's land. Ever since we have a change in sign this occurs between it goes from negative to positive that because that negative three extra cool negative Route three and then again happens our job. But there's one positive to negative zero, and then again every three and goes from negative fathers. We have three inflection point. No, and that is all.

Fahad P.