Showing posts with label 3C MAT. Show all posts
Showing posts with label 3C MAT. Show all posts

## Tuesday, April 3, 2012

### CAS calculator and differentiation

There are a host of ways to find the first derivative on the CAS calculator and to solve calculus problems quickly.  Sometimes I think exam writers are well behind what these calculators can do and fail to understand how trivial some problems have become.

a) Find y at x=3 for y = 2x^2 + 2x + 2

The "|" is important between the equation y = 2x^2 + 2x + 2 and x=3

Let's now find x if we know y.  This is a little harder as we have to solve the equation.

b) Find x at y=14 for y = 2x^2 + 2x + 2 The easiest way to do this is to type 14 = 2x^2 + 2x + 2, highlight it using your stylus (this is important!) and then go

Note that it find both possible solutions (unlike using numsolve with the incorrect range specified)

Let's find the first derivative.  For this use the 2D template in the soft keyboard Go keyboard -> 2D -> Calc ->

c) Find the 1st derivative of y=2x^2 + 2x + 2

Note that I removed the "y=" this time.  I differentiated the expression on purpose as it makes the next part easier.

Finding the 1st derivative/gradient/instantaneous rate of change at a point is also easy.

d) Find the 1st derivative of y = 2x^2 + 2x + 2 at x = 3

As you can see, it is a mix between c) and a)

Last but not least we can find a point for a particular gradient.

e) Find x at y' = 14 for y = 2x^2 + 2x + 2 To find y itself we could repeat a)

It's very much a case of thinking what you need and then finding it.  As you can see it can all be done with one line on a CAS calculator, things that would take multiple steps on paper.  TanLine is also a useful function that can be investigated and used to quickly find tangents.

Viola!

## Saturday, May 22, 2010

### Functions and the Casio Calculator fog(x)

When finding fog(x) the CAS calculator does a good job of simplifying algebraic steps, often to the point of making traditional questions trivial in the calculator section.

I'll start by defining a function f(x) =2x+1 and g(x)=x^2 and attempting to find fof(x), fog(x) and the inverse of (fog(x)).

To define a function f(x) go to the Main window and select

Interactive->Define

A window will appear

Enter the function name (eg. f)
Variable (eg. x)
Expression (eg. 2x+1)

Define f(x)=2(x)+1
done

Repeat:

Interactive->Define
Enter the function name (eg. g)
Variable (eg. x)
Expression (eg. x^2)

Define g(x)=x^2
done

To find fof(x)

In the main window
Action->Transformation->simplify
f(f(x)) (using the soft keyboard)
press exe

The main window should say
simplify(f(f(x))
4x+3

To find fog(x)

In the main window
transformation->simplify
f(g(x)) (using the soft keyboard)
press exe

The main window should say
simplify(f(g(x))
2x^2+1

To find inverse of fog(x)

In the main window
action->transformation->simplify
action->assistant->invert
y=f(g(x)),x))) (using the soft keyboard= don't forget the "y=")
press exe

The main window should say
simplify(invert(solve(y=f(g(x)),x)))

Here is a link to other CAS calculator posts.

### Inverse Functions and the Casio classpad 330

The 3CD course requires knowledge of domain, range, co-domain and inverse functions. The classpad can handle these in a number of ways.

The most obvious way is in graph mode (Menu-> graph tab). Set up a graph (say y=(x+1)/(x+2), and visually examine it to find the domain and range. To find the inverse, click on the graph and select Inverse (Analysis->Sketch->Inverse). The equation of the inverse can be found by selecting the inverse graph and examining the equation bar at the base of the screen. Be careful with this method, as the resultant inverse graph is not written in "y=" form.

The less obvious way to do it is in the main window. I have not found an "inverse" function yet, but the following is a workaround to find the inverse.
Go
Action->Transformation->Simplify (to simplify the resultant equation)
Action->Assistant->Invert (to swap the x & y variables around)
Action->Advanced->Solve (makes x the subject of the equation)
Enter the function
Press ",x)))" using the soft keyboard (to make x the subject of the new equation)

It would look something like this when finished:

simplify(invert(solve(y=((x+2)/3,x)))

In a way I prefer the main window method as it mirrors the algebraic method. I think students need to really understand the parts of an equation to effectively find the domain and range. I explicitly draw students attention to critical information such as the inability of the function to equal zero or where the function is undefined

a) Look for possible values of x where y=1/0 will occur (eg. {x≠-1} for y=1/(x+1) ).
b) If it is not possible for the numerator to be zero (eg. {y≠0} for y=2/(x+1))

If the range is not obvious it is often easier to examine the domain of the inverse of the function.

Here is a link to other CAS calculator posts.

Purplemath has a good explanation of inverse functions.

## Friday, March 5, 2010

### Absolute value

I spent a fair bit of time thinking about absolute value problems in the form |x+a| - |x-b| = c. Many students were struggling with visualising what these functions actually look like. What was happening when we try and solve them?

For example:
|x+5| - |x - 2| = 6
How could I display this equation graphically to give students an understanding of the underlying algebra to solve it?

I tried graphing y = |x+5| - |x - 2| and y=6 to find the intersection but was unsatisfied with the result as y = |x+5| - |x - 2| is not something easily tied to the absolute value concept or 'v' shaped absolute value graphs.
I was eventually satisfied with graphing y= |x+5| and y = |x-2| and then examining each part of the graph until I found a section of the graph that was 6 units apart.

For those wondering how to put it into a graphics calculator while exploring the concept

Go Menu -> Graph & Tab
Edit -> Clear All -> ok
at "Y1:" ->Softkeyboard->mth tab->select 'x'->type "x+5)" (it will change from abs(x+5) to |x+5|)
at "Y2:"->select 'x'->type "x-2)"
ensure that the boxes next to "Y1" and "Y2" are ticked

Now the temptation is to assume the answer is the intersection point.

but if we look at the equation |x+5| - |x - 2| = 6, it is asking "for what value of x is the value of |x+5| (the dotted line) subtract the value of |x-2| (the solid line) equal to 6". When is the gap between the two functions +6.

We can ignore values of x<= -5 as y=|x+5| is below y=|x-2| and the subtraction will only give negative values (we are looking for a gap of +6 which is a positive value).

We can also ignore values up to the intersection point as this also will only result in negative values.
The next place I looked is at x>=2 as the gap is constant and positive after this point (both functions have the same gradient).
at x=2, |x+5| is equal to 7 and |x-2| is equal to 0. |2+5| - |2-2| = 7. We can ignore values where x>=2 as the answer is not +6.

In fact the only possible solution has to lie between the intersection point (x~-1.5) and 2 and is probably closer to 2.
For y=|x+5| all values are positive between -1.5 < x < 2
For y=|x-2| all values are negative between -1.5 < x < 2
To ensure positive values for x-2 in the range -1.5< x < 2 we need to take the negative of (x-2) when solving the equation |x+5| - |x - 2| = 6.

x+5 - (-(x-2)) = 6
2x+3 = 6
x=1.5