## Monday, November 17, 2008

### Exponential Project Tips

As some questions are asked more regularly, I thought I'd provide some general discussion here.

(1) exp is the name of the function, just as sin and cos are names of functions. From calculus, you learn that sin'=cos and cos'=-sin.  This step shows that exp_b' = ln b * exp_b. (That is, it leaves the function alone except for a constant multiple. (But be careful where the chain rule is needed!)

(2) You do not need to use the limit definition (epsilons and deltas).  Instead, for perhaps the easiest solution, you should think about how to finish the statement:
lim b^x = lim [(b^x-1)/x ... ]
That is, if you start with (b^x-1)/x, what do you do to that expression to leave only b^x.  Then use elementary limit rules to compute your resulting limit.

(3) One method is to use the method of substitution for limits (change of variables) and then use an identity for the function so that the result of Step 2 is applied --- this method mimics what is done to show that sin x is continuous everywhere.  A second method is to use a general theorem that makes continuity an obvious conclusion of the results from Step 1.

(4) You must start with a statement like:
ln (1/b) = lim_{x → 0} [(1/b)^x - 1]/x
There are two easy approaches: (1) Find a common denominator to rewrite this as a simple fraction before continuing or (2) Think of (1/b)^x as b to some appropriate power and then use a limit substitution.

(5) Since you do not know the derivative of ln x, it is incorrect to use the Mean Value Theorem applied to the logarithm.  Instead, you should apply the MVT to the function exp_b(x) on an interval so that b^a and b^b are incredibly easy and where it is clear which value is larger (so that you know if the average rate of change is positive or negative).  You may use the fact that b^x is positive for all values of x.

(6) The function fb(x) is a linear function. You should write it in slope-intercept form (e.g., mx+b).

(7) Do not attempt to solve the equation fb(x) = exp_b(x).  There is one obvious solution from the definition: x=0.  But the formulas themselves do not explain where there would not be more solutions.  Instead, you should define a function (perhaps g) so that
g(x) = exp_b(x) - fb(x).
You know that g(0) = 0.  You need to show that g(x)>0 for all x ≠ 0.  My hint suggested Rolle's theorem, but I have since found that the Mean Value Theorem helps even more.  Use the Mean Value Theorem to show that for x>0, the average rate of change between 0 and x must be positive.  What about x<0? x="0?">

(8) You may not use a limit form of the type b. You may take a limit of the function fb(x) because that is of a form we know how to work with. Then you should use the result of (7) to conclude what the limit of exp_b(x) must be.

(9) and (10) put all of the previous steps together to perform analysis similar to Sections 4.2 and 4.4 to understand the graph.