sketch curve from function and plot derivative on curve

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deano2727

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Hi all,

I have a math exam on Tuesday and I am pretty much set when it comes to the exam papers, with one of the exceptions being the style of question shown below:

A function is defined as y=(x+3)(x+5)(x+8). Sketch this curve on graph paper showing the 3 points where it crosses the X axis (between the x values of -1 and -9).
Click to expand...

Not really sure how to approach this one. I tried getting the points by going in to my calculator and going to table mode and putting the function in to f(x) and then starting at -1 and ending at -9 in steps of 1. This just kept giving 'MATH ERROR'.

The next part of the questions asks to :

Find the derivative of y=(x+3)(x+5)(x+8) and plot on the same graph paper as part (a).
Click to expand...

I am unsure how to go about it after getting the derivative of the function, which I calculated to be 3x^2+32x+79. I am unsure how to get the points from this derivative.

Could someone be so kind to walk me through this question from start to end. Thanks!
 
Let's start with the places where the function crosses the \(x\)-axis, that is, where the function has a value of zero:

[MATH]f(x)=(x+3)(x+5)(x+8)[/MATH]
Since zero times any finite value is zero, we can simply find those places where the factors are zero. Equate all 3 factors to zero and solve for \(x\)...what are the 3 zeroes of the function?
 
To follow up...

We readily see that the zeroes of the function are:

[MATH]x\in\{-8,-5,-3\}[/MATH]
And we see that on:

[MATH](-\infty,-8)\implies f(x)<0[/MATH]
[MATH](-8,-5)\implies f(x)>0[/MATH]
[MATH](-5,-3)\implies f(x)<0[/MATH]
[MATH](-3,\infty)\implies f(x)>0[/MATH]
This alone is enough for us to make a reasonable sketch of the graph of \(f(x)\), but let's now look at the derivative, which you correctly computed as:

[MATH]f'(x)=3x^2+32x+79=3\left(x+\frac{16}{3}\right)^2-\frac{19}{3}[/MATH]
Application of the quadratic formula reveals the roots of this derivative to be:

[MATH]x=\frac{-16\pm\sqrt{19}}{3}[/MATH]
Knowing the vertex and roots is all we need to make a good sketch of the quadratic derivative.

We thus know that \(f(x)\) has turning points at:

[MATH]\left(\frac{-16\pm\sqrt{19}}{3},f\left(\frac{-16\pm\sqrt{19}}{3}\right)\right)=\left(\frac{-16\pm\sqrt{19}}{3},\frac{2}{27}(28\mp19\sqrt{19})\right)[/MATH]
Here is a plot of the given cubic function, with roots and turning points labeled, and the resulting quadratic derivative with dashed vertical lines at its roots:

fmh_0068.png

We can see that where the derivative is positive the cubic is increasing, and where the derivative is negative, the cubic is decreasing.
 
Thanks for the very detailed response, Mark. I appreciate you taking the time. I have figured out how to get the points on my calculator now.

The only part I am struggling with is on plotting the derivative of the function on the graph. I am unsure which metrics to use and how to get them. All I can get myself from that, is that you draw dashed lines down vertically from the turning points and the curve of the derivative will intersect the X axis here. How was the turning point of the derivative curve calculated? Sorry for the confusion.
 
The \(x\)-coordinates of the turning points come from the roots of the derivative. These are our critical values. Then we can use these critical values as inputs to the original function to determine exactly where the turning points are.
 
so the values of the derivative we would use to get our data is A=3 B=32 and C=79, is that correct? We can use the quadratic formula to get the first (X) value of each point and then sub that into the original first derivative to get the second (Y) value. That will give max and min turning points for the darker colored curve. What about the turning point of the lighter colored curve? That's the one I am unsure on how to plot. I can see it intersect the x axis at -6.786 and -3.88, but how do I determine the turning point of the lighter colored curve? I know how to get the turning points for the darker colored curve, but not the lighter one.

I am talking about the turning point marked below.



Sorry again, I am sure I am missing something stupid here.
 
deano2727 said:
so the values of the derivative we would use to get our data is A=3 B=32 and C=79, is that correct?
Click to expand...

Yes, those are the parameters we can use in the quadratic formula to get the roots of the quadratic derivative.

deano2727 said:
We can use the quadratic formula to get the first (X) value of each point and then sub that into the original first derivative to get the second (Y) value.
Click to expand...

Once we have the roots of the first derivative, we can use those in the original cubic function to get the coordinates of the turning points.

deano2727 said:
That will give max and min turning points for the darker colored curve. What about the turning point of the lighter colored curve? That's the one I am unsure on how to plot. I can see it intersect the x axis at -6.786 and -3.88, but how do I determine the turning point of the lighter colored curve? I know how to get the turning points for the darker colored curve, but not the lighter one.

I am talking about the turning point marked below.



Sorry again, I am sure I am missing something stupid here.
Click to expand...

The turning point of the quadratic curve, the first derivative of the original cubic function, is simply the vertex, which is why I expressed it in vertex form. However, we could find the derivative of the first derivative (the second derivative of the original function) to determine the turning point as well. But, because it is quadratic, we can use other methods, such as writing it in vertex form, or using the axis of symmetry, etc.

Using the vertex form I gave, we see the vertex of this quadratic is at:

[MATH]\left(-\frac{16}{3},-\frac{19}{3}\right)[/MATH]
Does all this make sense?
 
I am not quite sure I understand where you are having difficulty.

To sketch the curve describing a differentiable function, you need to plot the y-intercept (if there is one), the x-intercepts (if there are any), local extrema (if there are any), and points of inflection (if there are any). You should also determine whether the function has any asymptotes and what they are. Because it is just a sketch, you can approximate one or more of those points if finding exact answers is difficult or impossible. You then draw a reasonably smooth curve through those points while not intersecting any asymptotes.

Is there anything obscure about that?

The first derivative can help with all of that except for finding the y-intercept. Where a function's first derivative is positive, what can we say about the function itself? It is rising. How about when its derivative is negative? If the function has a local extremum at x = p, what will the value of the first derivative be? Zero right? So that will be an x-intercept of the derivative. (One of the points of this exercise is to make concrete that the derivative is a function in its own right.) Of course, not all x-intercepts of the first derivative are necessarily local extrema. You have to take into account the second derivative. How about inflection points? Well, the second derivative will be zero there (assuming the first derivative is itself differentiable). Where do we know the second derivative will be zero? At the local extrema of the first derivative (if there are any).

Any questions about that?

As for finding the local extrema (turning points) of the first derivative, just take its derivative and find out where it is zero.

[MATH]f'(x) = 3x^2 + 32x + 79 \implies f''(x) = 6x + 32.[/MATH]
[MATH]f''(x) = 0 \implies 6x = -\ 32 \implies x = - \ \dfrac{16}{3} \text { and}[/MATH]
[MATH]f' \left ( -\ \dfrac{16}{3} \right ) = \dfrac{3 * 256}{9} - \dfrac{ 512}{3} + 79 = 79 - \dfrac{256}{3} = 79 - \left (85 + \dfrac{1}{3} \right ) = - \ \dfrac{19}{3}.[/MATH]
Is that where you are lost?
 
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Thanks to both of you for taking the time to reply, I appreciate it.

I am teaching myself math after a long absence from it (and always being poor at it in the past), so some concepts take a little longer for me to understand. I have my own ways of learning and don't know some of the terms and calculations styles, so some of the calculation formats you guys have provided are a bit above my head.

So from what I gather about the above, to get the turning point of the quadratic curve, you get the second derivative which is 6x+32. Why does that positive 32 become a negative 32? I then see you divide -32 by 6 which returns -5.33.

Looking at the calculations for the Y value, I have no idea how you get to - 19/3.

Sorry again for the frustration I am no doubt causing in my confusion.
 
... Why does that positive 32 become a negative 32?
Click to expand...

f"(x) = 6x + 32 = 0

6x + 32 = 0

6x = -32

x = - 32/6

I then see you divide -32 by 6 which returns -5.33.

Looking at the calculations for the Y value, I have no idea how you get to - 19/3.

y = (x + 3)(x +5)(x + 8)

y = (-32/6 + 3) * (-32/6 + 5) * (-32/6 + 8) = ?
 
deano2727 said:
Thanks to both of you for taking the time to reply, I appreciate it.

I am teaching myself math after a long absence from it (and always being poor at it in the past), so some concepts take a little longer for me to understand. I have my own ways of learning and don't know some of the terms and calculations styles, so some of the calculation formats you guys have provided are a bit above my head.

So from what I gather about the above, to get the turning point of the quadratic curve, you get the second derivative which is 6x+32. Why does that positive 32 become a negative 32? I then see you divide -32 by 6 which returns -5.33.

Looking at the calculations for the Y value, I have no idea how you get to - 19/3.

Sorry again for the frustration I am no doubt causing in my confusion.
Click to expand...
So long as we can see you trying, we do not get frustrated.

Couple of points.

You approximated 16/3 by 5.33. It is a BAD idea in algebra and calculus to use approximations except sometimes at the very end. Here is a very brief explanation on why (the focus in the citation is on irrationals, but it applies as well to rationals with no finite decimal expansion).

How to approach this?

Here is the problem: I really couldn't come at this in any way. There seems to be not enough information. What I finally did was try to find the radius as a function of ....I'm not even sure of what but I called it X. This is what I tried: This came to nothing. It was a hard way tautology...
www.freemathhelp.com

The second derivative of a function is simply the first derivative of the function's first derivative. Just as you can find out a lot about a function from its first derivative, so can you find out a lot about the first derivative from the second derivative. Derivatives are functions. If a derivitive is a differentiable function, you can calculate and use the derivative's derivative just like any other differentiable function.

Finally, I was answering about how I found the coordinates of the specific point that you had indicated with a red arrow. That is not a point on the cubic function, but a point on the cubic's derivative, which is a quadratic. It is not just any point: it is the quadratic's turning point (local extremum). I found the x-coordinate by setting the quadratic's first derivative (which is the cubic's second derivative) to zero and solving for x. (Technically, I should have checked to make sure the quadratic's second derivative (the cubic's third derivative) was not also zero at that value of x, but that is obvious by inspection.)

NOW, to find out the other coordinate of the turning point of the quadratic, I must insert the value of x where the derivative of the quadratic is zero into the quadratic, not the cubic. It is important to remember that you asked about a point on the quadratic rather than the cubic. I hope this helps.
 
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Subhotosh Khan said:
f"(x) = 6x + 32 = 0

6x + 32 = 0

6x = -32

x = - 32/6

I then see you divide -32 by 6 which returns -5.33.

Looking at the calculations for the Y value, I have no idea how you get to - 19/3.

y = (x + 3)(x +5)(x + 8)

y = (-32/6 + 3) * (-32/6 + 5) * (-32/6 + 8) = ?
Click to expand...
My dear SK

The point the coordinates of which are being calculated is a point on the function's first derivative rather than on the function itself. So -16/3 must be inserted into the quadratic rather than the cubic. That will be 3^3 minutes in the corner.

:)
 
JeffM said:
My dear SK

The point the coordinates of which are being calculated is a point on the function's first derivative rather than on the function itself. So -16/3 must be inserted into the quadratic rather than the cubic. That will be 3^3 minutes in the corner.

:)
Click to expand...
So just to confirm, that would be:

y= 3(-16/3 )^2+32(-16/3 )+79
y= -6.3333

edited as I again approximated.
 
Thanks for all the help guys! You've given me the potential to earn an extra few marks on Tuesday.
 
JeffM said:
My dear SK

The point the coordinates of which are being calculated is a point on the function's first derivative rather than on the function itself. So -16/3 must be inserted into the quadratic rather than the cubic. That will be 3^3 minutes in the corner.

:)
Click to expand...
You are correct and I am in corner - but I am missing my buddy DB here....

Only feeble excuse - I lost site of the actual "find", flipping back-and-forth between screens.
 
Subhotosh Khan said:
You are correct and I am in corner - but I am missing my buddy DB here....

Only feeble excuse - I lost site of the actual "find", flipping back-and-forth between screens.
Click to expand...
Yes, we all miss him. He was mon ami.

The flipping back and forth makes sense to me. We are dropping you to 3^2.
 
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