Drawing derivative function. (2 Viewers)

asdfqwerty

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Say in questions where you are given a basic graph of a function, and you are told to draw its derivative, or given a derivative function and asked to draw the original function, how do you work it out?

I'm currently using Terry Lee textbook (3U). I have spent over almost 1.5 hours trying to understand how to draw the derivative function of a graph and I just can't comprehend it. I understand a few portions, but then a few minutes later, I get confused because the answers contradict with how I interpret how to work the questions out.


I don't have a scanner so unfortunately I can't scan some of the questions in the textbook, but can somebody generally explain to me how to draw derivative functions of graphs and how to interpret them?

Thanks.
 

deswa1

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There are a few ways to do this. The way I would do it is as follows. If this doesn't make sense, I'll try again:
1. Find stationary points on the graph (either by sight or by differentiating if you have the equation). These points are where dy/dx=0 so they are the x intercepts of the graph.
2. Look at where the curve is decreasing and increasing. (Where it is decreasing, dy/dx is negative and hence below the x axis etc.)
3. If the question requires more detail, look at the concavity. If the graph is concave up, then dy/dx is increasing at an increasing rate so the derivative function is increasing.
4. Connect the dots using this information.

For the derivative->orginial just do the opposite
Hope I didn't make any silly mistakes there (someone else might want to check, I'm too tired now) :)
 

Bored_of_HSC

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It's easy once you get your head into the correct mindset. It's simply using knowledge of what occurs when the derivative is = to or >/< than 0. You probably understand this but can't visualise or transpose it into drawing the other graph.

I cbf going through it all because i know you probably understand all these relationships. Just keep at it and most importantly, THINK about every step that you do.

NOTE: A good way to do these questions is to draw both graphs with one underneath the other. This makes comparisons easier (as you have the x-values of both graphs in similar places).
 

SpiralFlex

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Sketching the gradient function

So, I will quickly run you through the theory.

To fully understand how to sketch graphs of f prime x, we need to understand what we are actually doing when we are finding the gradient function. So let us begin this investigation!

We know the gradient function is a method we use to measure the gradient of the tangent at a particular coordinate. So we will now utilised this powerful tool and sketch A POSSIBLE curve for f prime x.

We will consider the following example. Over here we have a curve of . Nasty examiners will not tell you the equation of this curve, so we can't directly fully use the beautiful art of Calculus to sketch the gradient function. But however, there is enough information to sketch a possible shape.




Our first reaction is we notice there is one minimum turning point and one maximum turning point. What would happen if I take a tangent at the corresponding coordinate?



The tangents at those points are in fact horizontal. What does that mean? It means they have a gradient of zip! Zero! :)

How do we use this fact? On our original graph our axes were . On our new graph our axes are now

When we take the derivative of a function and graph it , our coordinate must stay the same. However our oordinate changes. In this case, the minimum and maximum points will become intercepts, since for maximum and minimum points (Stationary points),

So we can now swipe them down onto our pretty coordinates. Like so. (Note: The read resembles the graph on the (Just imagine it!) Sorry I had to do it on the same diagram for illustration.



Our graph is now divided into three regions! We must now consider each one. I will name the regions. A, B, C.




First letter of the alphabet goes first, Region A.



Over in this region our gradient is always positive except at the top, we established that it was zero, so I will put little '+' s to indicate that, if you don't believe me, we will investigate this further by drawing tangents.

We're still in Region A.



I have rated "arbitrary numbers" for the gradients to rate their steepness to illustrate my point. Clearly from Left to Right, our gradients are decreases. This means on our graph of from Left to Right values will decrease. In other words, our y coordinates will be decreasing as we go from Left to Right.

Or we can say as we go from Right to Left in that region, our y coordinates will tend towards infinity. So drawing our graph in this region, we have something that looks like this.





Second, we will consider Region B

Again for this region we notice the gradient is always negative. Let's arbitrarily rate the gradients again!



Remember these numbers are arbitrary!



This section we actually notice something very interesting, the gradients sort of get smaller, so the y coordinates of the graph f'(x) get smaller, but after that we hit a black spot. This spot is known as the point of inflexion. Then afterwards, we sort of rise back up again, and our gradients (y coordinates of f'(x) are getting larger.)

At the inflexion, our f'(x) graph will have a stationary point (minimum!) on the same x coordinates, but not necessarily the same y coordinates.

The inflexion is the lowest y coordinate of f'(x) we can get by inspecting the gradients! So this confirms it is a minimum!

So drawing this would look like.




Lastly, Region C

We can obviously see throughout this region there are all positive gradients, so the graph will ALWAYS be above the x axis in this region.




Now, examining their gradient with arbitrary values!




Ah, as we go from Left to Right, the gradient is increasing, therefore our y coordinates of f'(x) are increasing too.

Finally your graph is now ready to be served and eaten!



RAWWWWRR

[Editing]

-Refining detail and adding more description.
 
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Bored_of_HSC

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Sketching the gradient function

So, I will quickly run you through the theory.

To fully understand how to sketch graphs of f prime x, we need to understand what we are actually doing when we are finding the gradient function. So let us begin this investigation!

[Editing]
Observe... a genius at work.
 

asdfqwerty

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Untitled.png

Okay, so in this graph that I drew, a and b are both maximums and x=0 is the minimum.
Between a and b, the gradient is going down to zero, then up again to b.
In the derivative graph:

Untitled.png

What I don't understand is how that is translated into the derivative graph? In the f'(x) graph, the gradient concaves up and then concaves up to b which to me doesn't really make sense and I can't process my mind around it.
 

SpiralFlex

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View attachment 24014

Okay, so in this graph that I drew, a and b are both maximums and x=0 is the minimum.
Between a and b, the gradient is going down to zero, then up again to b.
In the derivative graph:

View attachment 24015

What I don't understand is how that is translated into the derivative graph? In the f'(x) graph, the gradient concaves up and then concaves up to b which to me doesn't really make sense and I can't process my mind around it.
Your region < a, always has a positive gradient. So the graph will increase to infinite (y coordinate) and will ALWAYS be above the x axis.

Your region a < x < 0 has a negative gradient. So the graph will always be below the x axis at that region.

Your region 0 < x < b again has a positive gradient. So the graph will always be above the x axis at that region.

Your region x > b again has a negative gradient. So the graph will be below the x axis tending towards -infinity.

I will clarify what I mean illustrated in the example above.
 
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Gigacube

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You may find that drawing the sign of the gradient on the graph may help you graph the derivative.

Untitled-1.jpg
 

asdfqwerty

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Hey Spiral, regarding Region B how do you know when the gradient changes its steepness? Are you meant to visibly see it on the graph without drawing the +'s and -'s?
 

SpiralFlex

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Hey Spiral, regarding Region B how do you know when the gradient changes its steepness? Are you meant to visibly see it on the graph without drawing the +'s and -'s?
I drew the + and - to help you visualise, obviously you can tell that the gradient is either positive or negative without doing the + and -

For region B, we know that the top line will have a higher gradient than the line next to it, we can visually tell. For example,



and are graphed. So we can see they have a gradient -1 and -2 respectively. It's like a visual aid my diagram.

Of course I have to point out the values are random! They just give an indication!
 
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SpiralFlex

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Sorry if I taking too long or am not good at explaining. It is quite difficult to explain over the internet especially questions with graphs! Also this is a brief guide. You will need to practice more and ask questions if you have difficulties!
 
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asdfqwerty

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Two more:

How would you do these functions, especially the one with the asymptote?

Untitled.png

Would these kind of graphs ever be tested in the exams? They seem very difficult.

P.S. Thank you so much for your help everyone! Rep'd you all :)
 

deswa1

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I'll try to explain the one with the aymptote-

Looking at the left hand side:
1. There is a maximum (stationary value) at around x=1 (just guessing as there is no scale). Here the derivative function is zero.
2. On the left hand side of this point, the curve is approaching the asymptote of what looks like y=x. That means that the derivative function is approaching the gradient of the aymptote. However the gradient is negative. Thus on the left hand side of the stationary point, the gradient function will approach -1.
3. Likewise, on the right hand side, the curve keeps geting closer and closer to the y axis. That means that the gradient keeps becoming steeper and steeper and will therefore go towards negative infinity.

If that doesn't make sense, I can upload a picture tomorrow so you can see what I did.

I didn't do that right hand side as that's pretty much the exact same thing just the opposite.

In regards to exams, I doubt you'd see something like that in a 3U test but I'm not that sure...
 

SpiralFlex

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The questions with regards to sketching graphs of derivatives involving asymptotes are usually found within trials of selective and private schools.
 

deswa1

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Hey Spiral, which program did you use earlier for those graphs? Also, nice explanation on how to do them, it was way easier to follow than my thing...
 

SpiralFlex

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Hey Spiral, which program did you use earlier for those graphs? Also, nice explanation on how to do them, it was way easier to follow than my thing...
Geogebra + Paint to edit. Very painful but do give it a go!

http://www.geogebra.org/cms/

Though I am looking for free software that will allow me to draw advanced 3D shapes in questions.
 
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cutemouse

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In regards to exams, I doubt you'd see something like that in a 3U test but I'm not that sure...
It was in a 2U or 3U HSC paper recently (can't remember which year but I think it was 2010) and these types of questions were my teacher's favourite in exams he'd set.
 

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