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# Match the graph of each function in (a)-(d) with the graph of its derivative in I-IV. Give reasons for your choices.

## a) IIb) IVc) Id) III

Limits

Derivatives

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##### Catherine R.

Missouri State University

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

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### Video Transcript

this private number three of the Stewart character. This eighth edition the section two point eight matched a graph of each function a through D with a graph of its derivative in one through four. Give reasons for your choices. S o. Here the grafts associated with this problem you have grabs a bcd which are functions F on DH graphs one, two, three and four. Here in red, which are a craft of the derivative of death, we need to match each graph of F with its served f prime. So beginning with part ay, here we have this shape of this function, and one of the easiest approaches is to identify the slopes that are zero those air ones that are easier to find because a slip of zero is easily found as a maximum or a minimum. And it could also be an inflection point. But it's clear when you drive the tension line to that point and the ten Jelena is horizontal. So this function here has a minimum, with the slope here of zero at here to the left of the Y axis and another slip of zero here to the right of the Y. axis. So what? This tells us that the function the derivative function will have a value of zero twice and on Lee, twice here to the left of the Y axis in here to the right deal axis. So if we look at the functions here below one, two, three, and four on ly, one of the functions satisfies this requirement. Ah, craft to graph one crosses thie XX is only one. So that doesn't count paragraph three cross of the X axis three times, including the origin. But at the origin, the soap is non zero. So this cannot be the answer. Grab him before does not crest across the X axis at all, meaning that it does not have any slopes of zero. Therefore, Kraft number two should be the best answer. And it is indeed the correct answer because the graph the slopes, the graph of the derivative is now only zero here to the left, to the right of why axes of the Y axis just like the function. But the maximum of the largest derivative occurs here at the origin. It's the steepest positive derivative and that degrees here, um, in graft number two, so we will say and connect craft, eh? Aircraft number two. And we matched the graph for the function f with derivative of prime. In that case, ah want a graph being Graff being seems to have a similar shape. Except the main difference here is that there are no derivatives or no slopes. Ah, with a slope of zero. So that's not easy to identify. Here we have three straight lines. One is a positive slopes. Tree line one is a negative slope. Straight line one another positive. So just a straight line. And where we should note, is that when it changes from a positive slope to negative slope or from the negative so to positive slip line, Um, the points here, our non differential, which means that there is no derivative at these points because you cannot draw attention, lane because these air corners, What that means is that there is no derivative know define derivative here at these points. So what we should take away from this function for B is that there's a positive slope and then it immediately changes to a negative slope and then immediately changes to pay positive school on because of the discontinuity In the derivative craft, we can rule out graphs one, two, and three Those air all continuous functions. The only function that satisfies his graph for because it's the only one with this continuity, ese. And we clearly see that the graph goes from positive derivatives to a negative, derivative or negative. So active father directive, so be matches with four. Okay, now on to graph see grafs. He has this shape and it's a little difficult. Identified the slopes here, but one that slope we do know definitely is. The slope here is zero had two Georgian at X equals zero at X equals zero. The tension line is horizontal, meaning that it's slope a zero. So the slope of the tenderloin zero and the function will cross the origin right here we see that to the left of the Y axis. The slopes at each of the points are negative, and they're increasingly negative to the left of the Y axis until it gets to zero. And to the right, just lips become positive. And one more thing we should know is that as we approach X equals infinity negative and committee tourist negative. Infinity and his exit purchase positive infinity. The function seems to level out approaching next slope of zero for the tender line. So we conclude that the function must cross the X axis. The derivative of this function was crossed X axis because of sloping zero x equals zero and that it should approach a slope of zero as Mexico sorting infinity and positive infinity and between the two remaining plants, plus one plus three. The only graft that satisfies this is ah, Kraft number one. So graft number two is not true. Three Graph number three's not true because three shows very large negative slope. Our negative derivative are very large, positive derivative Ah, in the negative extraction and a very large negative derivative in the positive extraction. And that's not what is happening here. Graft Number one shows correctly Ah that as X goes towards infinity, the slope that orbited function approaches zero and his expression they could infinity this slope function or the derivative a person's zero. And at the origin, the slope is zero, just like it said here and then we have negative slopes to the left of the Y axis and positive slips to the right of the wax is so grab. See matches with Raph number one. Finally, by process of elimination Graff ti matches of the Graph three and we can discuss exactly why that's true. Here we have this shape, we can identify three clear um, critical points. There's two local Maximus here where the tension line is horizontal. Here, to the left and to the right of the wax is the slope would be zero and then here at the origin the there's a local minimum and that slope is also zero of that tangent line. So it'll cross the our axes X axis three times and this is the only ground for that occurs Paragraphs number three and then we see that as X goes towards negative infinity. This function has a very large positive slope and his ex goes towards positive infinity. We see that this function has an increasingly negative so and that is also consistent with this graph. Number three

#### Topics

Limits

Derivatives

##### Catherine R.

Missouri State University

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

Lectures

Join Bootcamp