show-that-fxx2b-xc-is-decreasing-on-left-infty-fracb2right-and-increasing-on-left-fracb2-inftyright

Name: show that fxx2b xc is decreasing on left infty fracb2right and increasing on left fracb2 inftyright
Uploaded: 2019-06-26
Duration: 2 min 3 s
Channel: Numerade
Description: Show that $f(x)=x^{2}+b x+c$ is decreasing on $\left(-\infty,-\frac{b}{2}\right)$ and increasing on $\left(-\frac{b}{2}, \infty\right)$

Question

Show that $f(x)=x^{2}+b x+c$ is decreasing on $\left(-\infty,-\frac{b}{2}\right)$ and increasing on $\left(-\frac{b}{2}, \infty\right)$

Michelle Ling · Accepted Answer

relax is equal to X squared plus B R X plus C is increasing when x is less than negative, be over to an increasing when X is greater than negative view or two. So what we can do here is that we know if a if the derivative of a function is positive, the original function is increasing and if it is negative, the original function is decreasing. So what we can do here is find Theodore effective of this function so we can use the power down rule here. So the derivative of X squared is two acts on a derivative of be access be and you derivative c as a constant. So it&#x27;s zero. So what we can do here is we want to find out when the derivative will be positive, because that&#x27;s when FX will be increasing. So we can say to expose B is greater than zero and then we can solve for X, and we see that X is greater, then be over too. Now we can find the interval where it is less than zero so that the original function is decreasing. So we get X is less than be over too and we see that thes two intervals that we get are the ones that we intended to prove and to be getting.

Michelle Ling · Answer

So we want to prove that this function here after that X equals X to the third plus a X squared plus B X C is always increasing under the condition that B is greater than a squared over three. The first thing we want to do is we know that when a derivative is positive, it&#x27;s original. Function is going to be increasing so we can use that. Two. We can prove that the derivative of a backs is always positive. First we want to take the derivative. So we have three x squared close to a eggs close B on the car phone, the druid of a constant. It&#x27;s just zero. And first we can factor out the three leaving us with three times X squared, plus 2/3 a X plus be over three and now we can actually could use a complete the square method. So if we have X plus eight over three squared on, we have be over three left over and then we have we want to subtract a squared over nine which came from completing the square. So now we see that this half of the expression equals zero when X equals negative a over three. So this is true across negative infinity to positive infinity. So what we want to prove now is that this half of the expression is greater than zero for this hole thing to be to always be positive so we can rewrite this as three B minus B squared over nine is greater than zero on. So now we can also buy nine on both sides. We get three B minus. A squared over is greater than zero on. Then move a sward to the opposite side and then divide by three. And we get B is greater than a squared over three, which is the expression that we were trying to prove in the beginning. So now you know that if this condition is true, then the derivative will always be greater than zero, which means the original function will always be in

Michelle Ling · Answer

missions for A and B that make this function after vax equals X cubed, post expo speed always increasing. So we know that if the derivative of a function is positive than the original function, there&#x27;s increasing. So what we want to do here is find the derivative of F of X and see what these two variables need to be for it to always be positive so we can use the power down will hear the derivative of X cute is three. Looks weird on the derivative of X is A and B is a constant, so it&#x27;s derivative zero. So what we can do now is set this greater than or equal to zero on. We know that three X squared is always positive so we can solve. So we have three hugs. Squared is greater than negative A. And already we can see that a must be positive, um, for it to you, for the negative eight to be less than three x squared because X squared three X squared will always be a positive number. And then, as for B B, doesn&#x27;t really have any condition. Be can be any number because it is simply a starting point. It is the that why intercept of the function and where the function starts on the Y axis doesn&#x27;t really matter for whether it is increasing or decreasing.

Jonathon Brumley · Answer

We&#x27;re trying to determine where this function is increasing and where it&#x27;s decreasing. We do need to locate the critical point. So let&#x27;s take the derivative, which is to a X plus B. Let&#x27;s set that equal to zero. Okay, so we set that equal to zero and solve for acts. We would first subtract the B and then we would divide by the to a So X is going to be negative, be over to a now. There&#x27;s a certain consideration we have to take here. We don&#x27;t know if that A is positive or negative, and it does make a difference in the problem. Ah, this is the graph of a parable. A. So it&#x27;s either gonna open up or it&#x27;s gonna open downward. Let&#x27;s assume that a is positive. If a is positive, if we throw on a value greater than negative, be over to a we could be filling in something like negative, be over to a plus one. And if we happen to actually fill that in for X here, Wait, actually, right here we end up with an answer of one for the derivative. So that&#x27;s a positive on the other side. If we fill in something for negative, be over to a minus one. It comes out to be a negative, so if a is a positive, then it is increasing from negative. Be over to a to infinity and that&#x27;s really the Vertex and it&#x27;s decreasing from negative infinity to negative be over to a. But if a happens to be a negative number, then the problem opened the other way. So the increasing and decreasing regions switch. It&#x27;s now decreasing from negative be over to a to infinity while increasing from negative infinity to negative B over two A.

Helen Latting · Answer

Okay, So we&#x27;re trying to prove that if the derivative of F is always less than zero, then f of X is a decree decreasing function, decrease function. Um and that would mean that if a is bigger than be then s of a is also bigger than asked of B. Um, and the first thing to remember when doing this proof is that a derivative is just a slope. So we can write f prime of X as effort B minus F of a over B minus A. And that has to be less than zero. So we multiply both sides by zero. Only by B might. I say I&#x27;m sorry. And we get that s of B minus F of a plus to be less than zero. Um, and then if you add f evaded both sides, you get f B is less than f of a and that is your proof

Yaw Asomani · Answer

we want to show that this function here takes on it value, Uh, a plus B over to for some value off eggs, right when I should have this function here is gonna take on this value at some point for some value of X. I was going to use the immediate value through right meeting value. See him. Which means that if you have ah, any number, right? Any function that has two points. Say those and that or any points necessarily any pointed hook it always dysfunction or anything of that like that. Uh, then the function takes any value between that range. Any value here between that range, the function contain kit. That is what got into me and values saying and it&#x27;s always gonna have a correspondent X value, right? Definitely. Eso Uh, no. We&#x27;re gonna find at a right if a so ever a here is a minus a sway, then a minus B squared plus a right. This is gonna be zeroed, isn&#x27;t that entire thing is gonna be zero, and then you just gonna have a equals eight right in an effort. It&#x27;s me or ever be ever be Is B minus a squared, then B minus B square Cosby. Now, this is also gonna be zero. Because the miners be here is zero. Right? So, uh, f of bees. Those are gonna be be So this is a range, right? Arrange a and B so A and B is also a range, right? This is a d&#x27;Alene, and this is the range. Great. So it and he&#x27;s also a range no. Since and bees a range, then the by the immediate value through right, the function can take any value between this range that I did just here and disorder the range. Right. The function can take any value here between this range which has a correspondent as value. Of course. Right? Yeah. So Ah, now. So this is our range, right? If I should contain any value between these now, Doug, loss of generality, right. Without loss of generality. Ah, assume a is lesson being it&#x27;s less than or equal. To be great. Assume is less than I go to be. Let me clean this pace in order to use it. I want other things to be fund these same page, right? Eso We&#x27;re assuming that A&#x27;s less and be Dan a right A. You can rewrite a as the same as, uh, ageless A over to is the same thing. A All right, You can write s April say, over to you this is gonna be to a over tuition. Still a right. And we knew that that is gonna be less than ah a a plus b opportune. Since a is less than me lesson or you go to be, then this is gonna be less than or equal to a plus B over to right, And then this a Please be reduced. Also gonna be less than or equal to b plus B over to, because it&#x27;s less and be right. And this beepers be over to is the same as may. Right? So you can see that, uh, here. This a here has a plus B over to this interval. A and B here has a plus B over to in between, right, So you have a lesson or you go to a plus B over to less than or equal to be. That is what we&#x27;re getting from this. A little algebra there were did here, right? So this day, Polsby over to exist between this A and B range right exists between the A and B range. Therefore, about an immediate value theorem. The function here is gonna take the function here is gonna take this one for some value of X. So there is this on X. Call it C. Right. Such that f at sea here is gonna, uh, match this one. Right. Fc is gonna take on his value. So by the immediate value theorem, the function here can take there&#x27;s value. Uh, given that X is see right?

Problem

Show that $f(x)=x^{3}-2 x^{2}+2 x$ is an increasi…

Problem 54 Medium Difficulty

Show that $f(x)=x^{2}+b x+c$ is decreasing on $\left(-\infty,-\frac{b}{2}\right)$ and increasing on $\left(-\frac{b}{2}, \infty\right)$

Answer

For $x>-\frac{b}{2},$ we have $f^{\prime}(x)>0,$ so $f$ is increasing on $\left(-\frac{b}{2}, \infty\right)$

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For $x>-\frac{b}{2},$ we have $f^{\prime}(x)>0,$ so $f$ is increasing on $\left(-\frac{b}{2}, \infty\right)$

Calculus for AP

Grace H.

Kayleah T.

Samuel H.

Michael J.

Precalculus Review - Intro

Functions on the Real Line - Overview

Jon Rogawski & Ray CannonCalculus for AP

Grace H.

Kayleah T.

Samuel H.

Michael J.

Jon Rogawski & Ray Cannon

Calculus for AP