We've
seen the 'Sine',
Now let's
feel the 'Func'tions...
I was a bit limited in the types of functions that I could choose for
this exploration.
I was forced to choose a function that I could remember
how to take the derivative and anti-derivative of.
I chose a third degree polynomial. I must admit it took me multiple
tries to find an equation with a decent picture to talk about.
Below is the graph of the function y=(1/3)x^3-(5x)
In the class that I teach at Tigard High School we discuss several pieces
of a function, including relative extrema (Highs/Lows), increasing/decreasing
intervals, positive/negative intervals, and zeroes.
In order to describe the different intervals you must find some key
points on the graph. Relative extrema and zeroes are those key points.
If you solve the equation for y=0 you will find that x=0, x=sqrt(15), and
x=sqrt(-15). Those points are pictured on the graph where the graph
itself crosses the x-axis.
You can see the high point and low point on the picture but it's not
easy to see exactly where they're located. This is where the Derivative
comes in real handy.
The derivative of a function desribes the funcion's rate of change at
a certain point. Essentially the slope of the function. The
derivative of my original function is y=x^2-(5). Below the two functions
are pictured together, the original function in Purple and the Derivative
in Red. You'll notice that as the original function's slope is decreasing
(from infinity to approx. -4), the derivative is also decreasing.
When the original function's slope is zero, or changing from a positive
to negative slope, the derivative is at a zero. So, if we're looking
for the relative max. or min. we can look to the zero of the derivative
to find the exact x-value for each. The derivative's zeroes are at
the sqrt of 5 and the opposite of the sqrt of 5. If I plug
those numbers in for my original equations I can find the relative extremas.
The points (2.23, -7.45) and (-2.23, 7.45) are the relative extrema.
Now, if we find the anti-derivative of the original function, then
all the same rules should apply for our original equation that applied
for the derivative that we graphed and talked about earlier. Below
the three functions are graphed. The two we talked about earlier
and the anti-derivative, y= (1/12)x^4-(5/2)x^2, in blue. You can
see, once again, that where our original function is at a zero, the anti-derivative
is at a relative minimum. Isn't that splendid...
Now let's go take a look at a couple of functions we have already explored...The
Sine and Cosine Functions
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