Untitled Document

Printer-friendly version

03 October 2003

Extreme Values of Functions

by George E. Hrabovsky, President, MAST

Where We Have Been

We have been exploring derivatives. Last time we developed the constant rule, the product rule, and the power rule.

The Same Problem in the Lab, For the Last Time

From last time we developed the expression

$FormBox[RowBox[{x/t, , =, , RowBox[{3 t^2, +, 68 t, , -, , RowBox[{127., C ... p; ], (1)}]}]}], TraditionalForm]$

You show this to your colleague and feel good that you answered his question. He smiles and asks, "So, what are the extreme values of your original function?"

You frown, "What do you mean?"

He shakes his head slightly, "I mean what are the maximum and minimum values that the function can take?"

The good feeling you had before vanishes. How do you proceed? You think about this for a while and it eventually occurs to you that if a function reaches a maximum and/or minimum value there will be no change above or below that value of the independent variable. The value of the derivative at such a point will have to be zero, since these are constants of the function under study. So, you set the derivative to zero,

$FormBox[RowBox[{x/t, , =, , RowBox[{3 t^2 + 68 t - 127, =, 0.}]}], TraditionalForm]$

Now you solve for . Since this is a quadratic we have

Of course, this has two possible values,

or, and . We can insert these values into our original function

to get,

$FormBox[RowBox[{x, , =, , RowBox[{RowBox[{5514.48, , and, , x}], =, RowBox[{-, RowBox[{2.28419*10^6, .}]}]}]}], TraditionalForm]$

We call these critical points and we will explore more about them next time.

The Product Rule

The second rule we have conjectured is called the product rule.

The Power Rule

The third rule we have conjectured is called the power rule.

The Math Challenge from Last Time

I posed the challenge of proving three f differentiation rules last time. The first was the constant rule. The argument here is trivial,

The second rule is the product rule,

&nbs ... )]/(Δ t) + Underscript[lim, Δ t  0][u(t + Δ t) v(t) - u(t) v(t)]/(Δ t)

The third rule is the power rule,

We can then use the definition of the binomial coefficient to make this shorter,

so we have,

/tx^n = Underscript[lim, Δ t  0] (x(t)^n + (n) x(t)^(n - 1) Δ ... 1 2 n

= U ... 1 2 n

The Math Challenge

How would you take the derivative of a quotient?

Math Resources for Derivatives

Online:

http://www.ma.utexas.edu/users/kawasaki/mathPages.dir/

This is a neat site. For a more challenging page look at this one:

http://www.mathpages.com/home/icalculu.htm

Both sites are very cool.

Books:

Richard A. Silverman, 1969, Modern Calculus and Analytic Geometry, MacMillan Company, New York (Dover Publications has reprinted this book with corrections in 2002). This has a rigorous, but readable chapter on derivatives.

Created by Mathematica (October 3, 2003)