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Question

Analyzing a Critical Number A differentiable function $f$ has one critical number at $x=5 .$ Identify the relative extrema of $f$ at the critical number when $f^{\prime}(4)=-2.5$ and
$f^{\prime}(6)=3$

Pagadala Kishore Reddy · Accepted Answer

all right. It was given that No, Your pass. Well, remember it. Exit was two friends. If there&#x27;s food equals minus put blind faith, which is a let down Jill. So the function is decreasing at X equals 24 And if they&#x27;re six equals 23 which is greater than zero. So the functioning of years increasing attacks in close to six here the same off first enemy too. Abdel function their profits. These change it from you bother to. So according to the I didn&#x27;t mean to guest, we can see that. Frank, come on. Yes, I&#x27;m afraid easier. Mukhin, we need more. Thank you for watching.

Sahil Patel · Answer

a cripple pointed to before, too. We know it&#x27;s increasing, and after two, we know it&#x27;s also increasing. There&#x27;s no critical points between one and two and doing three. That means that there&#x27;s no extreme at two, it&#x27;s

Elizabeth Treanor · Answer

Okay, Question 29 is asking us to find the critical points of the function. F of X equals five x squared plus four X and critical points is when the derivative equal zero or where the derivative doesn&#x27;t exist. In this case, since the function has a continuous domain from negative infinity to positive infinity and it&#x27;s derivative also has a continuous demean from negative affinity to positive. Infinity doesn&#x27;t exist. Isn&#x27;t going toe apply in this example. So instead, we&#x27;re just going to look for where the derivative equal zero. And if you&#x27;re looking at the graph of a function, the derivative is going to equal zero at extreme points. So at a maximum at a minimum or at a point on the graph that has a curve similar to the cubic function and all of those points Ah, have something in common. They all have a slope of zero, so these critical points are always going to occur when the derivative or the slope is zero. Looking at the graph of five X squared plus four X that only happens once, and it&#x27;s happening at this minimum point here. But I&#x27;m during a tangent line through right now. So that is the only critical point on this graph. Looking at ah different graph really quick. This has two critical points. So I&#x27;m doing a tangent line through one now at this minimum, and the 2nd 1 is through that curb I talked about That looks like a cubic function. Okay, but back toe this example. So we&#x27;re looking for one critical point here, So let&#x27;s find the derivative now. So the function itself is five x squared plus four x. So the derivative of this function gonna do two times five and get 10 X. And now the exploding is gone. Plus four on the variable for that is also gone. We&#x27;re gonna set this derivative equal to zero and now solve for X. So subtract four on both sides to buy buy 10 on both sides. You get X equals negative for over 10 or X equals negative 2/5. Now, that is just the X value of this critical point. So to get the y value, we&#x27;re gonna substitute it back into our original function. And that is f of X equals five X squared plus four x. So instead, we&#x27;re gonna now be finding f of negative 2/5, which equals five times negative to over five squared, plus four times negative. 2/5. We get five times negative to over five. Squared is positive for over 25. Four times negative to his negative, eh? It&#x27;s so minus 8/5, and we&#x27;re gonna continue to simplify this. So five goes into five once and five goes into 25 5 times. So we get 4/5 minus 8/5, which equals negative 4/5. So f of negative 2/5 equals Negative. 4/5. Meaning that this critical point. But we were looking for that. I have now Kraft in black. Above is the point negative to over five comma, negative for over five.

Carson Merrill · Answer

he received two areas of intersection. Um, this would be a relative maximum because the slope is positive and then becomes negative. This would be a relative minimum because the slope is negative and then becomes positive.

Carson Merrill · Answer

here&#x27;s Young&#x27;s graph that zero because it has a negative slope and then it stays negative. This would just be it wouldn&#x27;t be a minimum or maximum. And then we see right here that there is actually ah, a minimum maximum because the slope is negative and then it becomes positive. That would be like slip being negative and then positive. So it&#x27;s gonna be a relative minimum. Um, at this point, it says 1.71 We can also do math to show that it is exactly, uh, the Cube Group of Five, and that would be a minimum.

Amrita Bhasin · Answer

Okay. We know that we can use the product. Roll off one GPAs FT one to find the derivative were fracturing out the two, as you can see. Okay, Now we have X minus form parentheses. Their first set, X Honest four equals to zero on the end, up with one of our critical numbers. X equals four. Next, we know that we can set seven x months. Eight equal to zero and we end up with X equals eight over seven as our second critical number. And then lastly, don&#x27;t forget that we have zero as our third critical number because of the denominator.

Problem

Analyzing a Critical Number A differentiable func…

Problem 71 Medium Difficulty

Analyzing a Critical Number A differentiable function $f$ has one critical number at $x=5 .$ Identify the relative extrema of $f$ at the critical number when $f^{\prime}(4)=-2.5$ and
$f^{\prime}(6)=3$

Answer

$(5, f(5))$ is a local minimum!

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$(5, f(5))$ is a local minimum!

Calculus of a Single Variable

Catherine R.

Heather Z.

Kayleah T.

Michael J.

Precalculus Review - Intro

Functions on the Real Line - Overview

Ron LarsonCalculus of a Single Variable

Catherine R.

Heather Z.

Kayleah T.

Michael J.

Ron Larson

Calculus of a Single Variable