💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here! # Use the given graph to estimate the value of each derivative. Then sketch the graph of $f'$:(a) $f'(-3)$(b) $f'(-2)$(c) $f'(-1)$(d) $f'(0)$(e) $f'(1)$(f) $f'(2)$(g) $f'(3)$

## It appears that $f$ is an odd function, so $f^{\prime}$ will be an even function-thatis, $f^{\prime}(-a)=f^{\prime}(a)$(a) $f^{\prime}(-3) \approx-0.2$(b) $f^{\prime}(-2) \approx 0$(c) $f^{\prime}(-1) \approx 1$(d) $f^{\prime}(0) \approx 2$(e) $f^{\prime}(1) \approx 1$(f) $f^{\prime}(2) \approx 0$(g) $f^{\prime}(3) \approx-0.2$

Limits

Derivatives

### Discussion

You must be signed in to discuss.
AM

Ava M.

October 11, 2021

is there all of centage questions answer in here?

##### Top Calculus 1 / AB Educators ##### Heather Z.

Oregon State University  ##### Samuel H.

University of Nottingham Lectures

Join Bootcamp

### Video Transcript

for this problem, we need to calculate some and derivatives at some given point and we are only given the graph for the function F. S. So what we are going to do is to calculate this globe of the tangent function at each of these points. And because that's the definition of a derivative in a given point. So what we are going to do in this first point, we have to calculate the derivative of the function in the position minus tree. So we see then -3 is in here and a tangerine lined, we just draw attention line and we just can see trying attention line. So with this change in mind we can see that we are given two points. The first point is in -8 and this corresponds to behave or white component of approximately -2 points sorry minus one mhm minus one point nine. And which is this that I went to mark quitting red color and the other point is this one here so we can see that this point is -1 in the x company and approximately mm nine is 2.3. So with two points we know that knowing two points of a function or of uh in this case the tangent line, we can know the slope by following. And this formula, the slope is equal to the final point that we're going to call it as put in here, Add effort. S two. S two -F of X one over. It's 2 -11. In this case we know that This first correspond to S one, these two as two. And these two to have the evaluation of the function effort as one and this effort adds to So substituting this in here, we will have minus 2.3 minus minus 1.9 over minus one minus minus dream. So this give us a value of -0.2. So approximately the Derivative of the function at the 0.-3 is equal to -0.2. So this is the first derivative that we can that we just have calculated now we want the derivative at the point minus two. So again we see the pointer is in here making a pointer and we just dropped touch in line to that point, we can see that drawing a tangent line to that point will produce and our son told mine. So this is very strange for work. We know that the slope of an horizontal line is zero. So we can just simply say that The derivative of the function in the 0.-2 is just zero because um it is equal to the the slope of the tangent function. And this lobe in the time of the tangent function in this case is equal to zero. So we will have that now for minus along. We see that disease in here. So we're gonna just drop there and want to try to draw various trade line and sarin try against here. Yeah, thanks. Just let's try again. A better one. Okay just that's gonna that's gonna live in like you know it's just it needs to be like a tangent life so you're gonna try better. Okay that's better. So we have to point the first one and we can see from the graph, am gonna mark it in red is approximately um their wine component is approximately minus one point fine. And the other one, the other point sit here could sit at this point in here on the market in here and this point is indeed at Uh the x component should be 0.5 And the White Continent is just -1. So from here we just again can obtain the slope. So the slope in this case we just know using the is formula and effort S. Two. Is this in here? So we'll put in here minus mm minus F. As one over adds to minus minus one. So this give us a value off, Give us the values of one. I'm sorry this give us a value off. Yes it gives values of one. And so that we conclude that from here we conclude that the derivative of this of this function and it's equal to -1 is equal to what now for F. Or the Derivative of the function at zero zero is here. And which is straw. A tangent line 22 22222 to to to to you self We have to point the first point is 00. Okay and the second point we could see that is let's try it better because it's not like a tangent line. It should be more. Okay lets okay, so kind of starting here we can see that if we're starting here passing here some bettors. Okay. Yeah, okay so in this and this case we have zero series. The first point and the second point is approximately in here is minus one in the expert and -2 in the white part. So the slope is simply again we use the Formula -2 0 This case And -1 0. So this gave us a slope of two. So we conclude that the derivative of the function at zero is equal to two. Yes, no for the following value for Derivative at the value is equal to one. It's in here the point. So we're gonna try to draw our best a tangent line. So let's see, we're gonna try starting here us in the point. So okay, so we have the point want one and in here one We're going to take these two points. So this thing here is one and the white part is the white company is 1.5. The second point is it's a little higher is a little higher of of zero point by so we're going to say that is It's 0.6 and the export of course it is indeed zero. So taking this slope, we know that Again. This is zero Point seats -1.5 over zero minus one. This give us a slow off. Yeah. Zero point no. So approximately the derivative at the .10.9 the net derivative is S equals two. So at s equal to as in the case of It's -2, we will have and horizontal tangent line. So we just conclude that this is equal to zero. The derivative at the point X equal to two. Now our F M. The liberties at at equal dream we will have this in here and which is try to draw our best tangent line. So like this. So we are gonna take these two points went in here and one in here. So we have the 0.3 uh 1.9 approx. And the point the second point it's one in the ads company and or the white part. All right Is approximately 2.3. We're going to say that is 2.3. So they can, the slope of this is gonna be um 2.3 -1.9 over one minus three. So In here we will have a value of -0.2 And just give us approximately a value for the derivative at the .3 of zero point. Now to graph this function, the the function, the f prime function or did everybody the derivative of the function F. S. We just need to just um first we are gonna like just copy all the information in here. The value of S. And the value of F. F. S. F. The derivative of S. That we just have obtained. So for s We have the values minus three minus two minus one. Zero. Want to and tree. And for for the Eppes to we're just have that were the first one. We hope saying 0.2 for minus 20. For minus one. Want Force zero with team two for one. uh 0.9 and four to uh huh. Or toothy room. And for three again wore zero my 0 Mm zero pointed. So we're just gonna put all this in the garage and we just catch the graph that we are we can obtain from just right. So for minus dream that is in here minus dream. The value of the function is zero. So it is very like approximately in there. We're going to put it in a red color to differentiate Now for minus student is in here -2. We have the real. So is here indeed. It's abscess. Um for -1. It is in here minus minus want it's one that is in here For zero. It's too four. one is 0.9. We are going to like well it's kind of here for two. It's zero against zero in here and for three is minus to So we have a teams, almost a very symmetric function. So if we just unify all these points, we obtain that this is something of this form. We we cannot be sure exactly how the function is, that it's it's going to be some kind of this form. So this is the graph of the function of the derivative of the function. Uh That's that we start with. Central University of Venezuela

#### Topics

Limits

Derivatives

##### Top Calculus 1 / AB Educators ##### Heather Z.

Oregon State University  ##### Samuel H.

University of Nottingham Lectures

Join Bootcamp