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# Use the given graph to estimate the value of each derivative. Then sketch the graph of $f'$:(a) $f'(0)$(b) $f'(1)$(c) $f'(2)$(d) $f'(3)$(e) $f'(4)$(f) $f'(5)$(g) $f'(6)$(h) $f'(7)$

## a) 6.7b) 0c) $f^{\prime}(2)=-1.5$d) -1e) -1f) $f^{\prime}(5)=-1 / 5$g) 0h) $f^{\prime}(7) \approx 0.2$

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this problem. Number two of the Stuart Calculus eighth edition section in two point eight. Use the given graft to estimate the value of each derivative, then sketched the graph of F crime. So this is a graph of F given as this certain shape. And this is a graph for this problem, and we're hoping to sketch a graph of F prime or the derivative of dysfunction. And we're going to do that by finding the derivative for each point from zero to seven. X equals zero tech sickle seven and then using each of the individual values on DH this plot here, tio, plot what our function will look like. Okay, so we begin with X equals zero. What is the slope? What is the derivative at the origin we look at our function on? We see that it has a pretty steep, positive soap. It goes approximately up little more than three. This is with attention. Lang would look like, uh so it's not to re over one the slope. It probably is more closer to four over one or five over one from as to how steep the slope of this tangent, Linus. So as an estimate. We're going to say that the rise over run is for for the slope at the origin. Carping at X equals one. What? It's a derivative. What is the tension? I ll look like tension in line at X equals one. It looks like this all right, touches the function at X equals one only at one place and the line is horizontal, meaning that it has a slope of zero at X equals one. Add X equals two. We take a look at where the function is at X equals two. It's right here. Let's say that we went down one and over one thats not quite a line that is tangent to the function. And so we go down to approximately and over one that seems to be a little too steep. So when they get one over one, it's not steep enough. Negative two or one a little too steep. How about halfway halfway between negative one Negative too. That seems more appropriate as attention line, uh, for this function. So half way we can make it a one or negative to three over to, and that will be our estimate for the Slope X equals two at X equals three. What is the derivative? What is this stuff of the attention line X equals three. This is the point here, crossing that Texas. And we see that if we draw attention, line on that it is closely associate ID approximately ah, equal to a slip of negative one. So that is our best estimate. We're going to go with a slip of negative one for At X equals three and X equals four. We look at where that point is. Two. Three, for this is the point X equals form. We're trying to estimate what potential and my look like we'd go down one and we'LL go over one thats not quite steep enough or it's a little too steep. So we go over to and this seems to be a bit more appropriate if we were to draw this tension line here. That seems to be a pretty appropriate tension. Lame at X equals four. And this slope here is down one over to a rise of negative one over two. That is a slope of the negative one house annexe Eagles fry. What is the derivative of F one, two, three, four, five This is the point for the function. There's a slope world. Well, tryingto as to me by going up one in going over one, two. Uh, it seems to be a good estimate. How about three? I think that is a little better. We know that the slope add five as it gets closer to the next value. I'm The slopes are going from steeper tonight. A Steve. So this seems to be a better estimate. Um, the sloping arise of negative on and a run of three or closer to that than negative one over to. So we're going to see that they're slow. Estimate at X equals five is negative. One over three at X equals two six. This is where the function is at X equals six. We see that it is it horizontal pendant horizontal line on the slope of zero. So six there's a minimum and the slope of the tangent line zero. So the dirt of zero and finally at X equals seven, one, two, three, four, five, six, seven. This is where the point is. All right, if we go down and then over one too steep for the soap over two Still seep three, four five at five. We imagine it a more approximate tangent. So this is what ah, run of five and a rise of naked one would look like. So this being a slow a rise of one. So this is a rise of one and a run of five. That is a slope of one over five or one fifth. And this seems to be more amore. Appropriate estimate for the tension. This liberal potential in an X equals seven. So we're going to go with one over five. Great. So finally, we're going to plant this function in this F prime function, and we're going to supply each of these points for X equals zero through X equals seven. Starting with X equals zero. The point. The value of the derivative is for the next point at X equals one. The value of the derivative is zero for X equals. So we're gonna draw a couple things here in a draw, make it one, continue the function a little lower, and then draw negative too. Okay, being this in mind, what is the next point at X equals two. The stop is thinking of three over to our negative one half that's approximately right here at X equals three. The value of the dirt of his negative one right about there at X Eagles before the value is approximately negative. One half right right there had X equals to find the value of the dirt is a personal maid of one third. So just a little higher closer to the X axis. An ex eagles to six, the value the dirt of zero. So now across the X axis. And Alex, he was the seventh valued. The jury was one fifty. So this is what the shape of the function will look like. Now we found thie. Sure, if it is at each of the points. And now we plotted them separately and we join them with the single smooth curve to estimate and sketch the parent behavior of this function of crime. So this is a sketch of the graph of Prime that we found it right from the individual irritants. Eight different points of the graph of F given here and that completes this problem

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