Lab 08: Graphs

0. Objective:

In this lab we use Mathematica to explore the different aspects of graphs.  To do this, we will use the following terminology:
1.  A function, f, is said to be increasing/decreasing on an interval I if the graph of f rises/falls when traced from left to right on the interval I.
2.  We say that the graph of the function, f, has a high point/low point at some number a if there is an (possibly small) open interval containing a such that the point (a ,f (a)) is the highest/lowest point in the graph of f over this open interval.
3.  We say the graph of the function, f, has a point of inflection at some number a if there is an (possibly small) open interval containing a such that at the point (a ,f (a)) the graph of f changes concavity.

1. Examples:

Consider the following function.

f[x_]=(x^3)-9*x+2

In order to better understand what range of values we should use to graph this function, we will first investigate the attributes described above for this function.

First the rise/fall of the graph.

cv=Solve[f'[x]==0,x]

We can graph the derivative to see where it is positive and where it is negative, thus telling us where the graph of f is increasing and where it is decreasing.  Because the zeros of the derivative are ±(Sqrt[3]) we need only look at an interval, like [-2, 2], that contains those values.

Plot[f'[x],{x,-2,2}]

Since the derivative is negative on the interval (-(Sqrt[3]), (Sqrt[3])), the graph of f is decreasing on this interval.  Like-wise, the graph of f is increasing on (-∞, -(Sqrt[3])) and ((Sqrt[3]), ∞).

Next, the high and low points.

We found the critical values above.  To find the high and low values we need to evaluate the function at these values.

N[f[x]/.cv]

Looking at these values (and the first derivative test), we can say that we have a high point at (-(Sqrt[3]), f (-(Sqrt[3]))) and a low point at ((Sqrt[3]), f ((Sqrt[3]))).

Next, let's look at the concavity and points of inflection.

Possible points of inflection are where the second derivative of the function is zero or undefined.

ppi=Solve[f''[x]==0]

If we graph the second derivative we get

Plot[f''[x],{x,-1,1}]

Looking at this graph, we note that the second derivative is negative on the interval (-∞, 0), so the graph of f is concave down on this interval.  The second derivative is positive, and the graph of f is concave up, on the interval (0, ∞).  Thus, we do have a point of inflection at (0, f (0)).  Note that we can evaluate the function at the possible points of inflection by the following.

f[x]/.ppi

Finally

The last of the interesting aspects of the graph of f would be the intercepts (note that there are no asymptotes for this function).  The y-intercept is the point of inflection.  To find the x-intercepts we could use the following command in Mathematica.

xint=Solve[f[x]==0,x]

This doesn’t look very nice, so let’s try doing a numerical approximation.

xint=NSolve[f[x]==0,x]

That’s much better.  To summarize, all of the interesting properties of the graph of f occur in a window of about [-4, 3]×[-9, 13].  So now we can graph f on that window to get:

Plot[f[x],{x,-4,3},PlotRange->{-9,13}]

Name:
PIN:

Task:

Consider the following function:

f(x)=((384-64 x-376 (x^2)+148 (x^3)+54 (x^4)-25 (x^5)-2 (x^6)+(x^7))/(2+3 x+(x^2)))

(1)  Find the critical values of this function.

(2)  Find the intervals on which the function is increasing.

(3) Find the intervals on which the function is decreasing.

(4)  Find the high and low points of this function.

(5)  Find the possible point of inflection.

(6)  Find the intervals where the function is concave up.

(7)  Find the intervals where the function is concave down.

(8) Find any points of inflection.

(9)  Find and classify any discontinuities (asymptotes, holes in the graph, ...).  Also find the intercepts.

(10)  Use the above information to graph the function in an appropriate window.  State the window and sketch the graph on this sheet.