Important grading issue: The following issue was brought to my attention by one student. On page 96 of the text, the derivative of a function f(x) at x = a is defined to be synonymous with the instantaneous rate of change of f(x) at x = a.
On the previous page, the instantaneous rate of change is defined as the limit of average rates of change, namely,
(inst. rate of change at x = a) equals
(lim as h -> 0) (1/h) (f(a+h) - f(a)).
On the test, I wanted this limit as the definition of derivative, but as the text does the definition in two stages, I think I have to allow the definition on page 96 as a proper answer on the midterm.
If you gave the definition on page 96 and were marked off for it, see your TA and get him/her to give you credit (4 points) for problem 2b.
However, the next time this question is asked, I insist on your giving the limit definition for the derivative,
f'(x) = (lim as h->0) (1/h) (f(x+h) - f(x)).
That is, I want an answer with content, not just a synonym.
I apologize for any stress and inconvenience this may have caused you. (In particular I apologize to the student who brought this to my attention -- at first I refused to consider a change of grading, but on further consideration, I decided he/she had a valid point.)
(1) Section 4.3, Exercises 1 through 8, 12 through 17, 21, 25, 28. (15 and 25 are presentation problems.)
(2) Aditional Graphing exercises: for each of the following functions, compute the first and second derivatives. Find out on what intervals the function is increasing/decreasing and on what intervals the function is concave up/ concave down. Calculate the local maxima and minima, and the points of inflection (points where the concavity changes). Consider what happens to the function for large positive and negative values of x. On the basis of all this information, draw a sketch of the function, with the several important points labelled -- i.e. the local maxima and minima, and the points of inflection. Finally, get Mathematica to make a plot of the function and compare the computer plot with your sketch.
a) f(x) = x^3 exp(-x). (Presentation exercise!) Even though the function seems a little complicated, it turns out to be easy to find the both critical points and the places where the second derivative is zero.
b) f(x) = (7x^2 -2)/(3x -10)
This one has a vertical asymptote and it has two critical points, one on each side of the vertical asymptote. Show that one of these is a local max and the other is a local min. Get Mathematica to plot the function for some distance on either side of the critical points to see clearly the behavior, for example,
f[x_] := (7x^2 -2)/(3x -10)
Plot[f[x], {x, -20, 20}]
(By the way, Mathematica code given in these assignment sheets can be cut and pasted into a Mathematica notebook for evaluation. You don't really have to re-type it.)
Also note that this function has a slant asymptote; that is the graph approaches a certain line of non-zero slope for large positive and negative x. To investigate this, try graphing the following line and the function together:
f[x_] := (7x^2 -2)/(3x -10)
g[x_] := (7/3) x
Needs["Graphics`Colors`"];
Plot[{f[x], g[x]}, {x, -30, 30}, PlotStyle-> {Red, Blue}]
Note that y = (7/3) x seemed like a good guess for the asymptotic line, since the function "behaves like" (7x^2)/(3x) = (7/3) x for large x. But from the graphs, you see that y = (7/3) x is not exactly the right asymptotic line. It appears to be parallel to the correct asymptotic line! How could you find the correct asymptotic line?
(3) An optimization problem: An Iowa farmer keeps free range chickens. The chickens do not swim. He has a straight river front on his property and 1000 ft of fencing. He decides to make an enclosure for the chickens, rectangular in shape, with one side formed by the river and the other three sides by the fencing.. He wishes to give the chickens as much area for movement as possible. What dimensions should he make his enclosure in order to maximize the area enclosed. Be careful to find the appropriate domain of your independent variable and to check the end points of the domain for a possible maximum.