L'Hospital's Rule, Min-Max (local & absolute)
              Curve Graphing, Newton's Method

                              Mathematica

Dr. Andrzej Korzeniowski
Department of Mathematics, UTA

Lab 4

L'Hospital's Rule

To find    x [Log [x + 5] - Log [x ]]  one may graph the expression first:

which suggests the limit at ∞ is 5.

To apply  L'Hospital's  rule notice that the expression can be written as   = with   at ∞ .

Consequently  the limit of the original expression is

Local Extrema

Critical  numbers = {x | f ' [x] = 0 or  f ' [x]  does not exist} constitute a key element for studying maxima or minima (extrema) of  f [x].   

Pick the real solutions only (use cut & paste )

To verify that f [x] achieves a local  max  at  x = .902652 with  max = 5.1841 and local min at x = 2.56238 with  
min = f [2.56238] = -105.931 one can utilize the second derivative test as follows:

which shows that f ''  is negative (as required) at x = .902652  and f '' is positive (as required) at x = 2.56238.

Another way to find the min or max of  f [x] is to use the built-in numerical algorithms as follows (provided that
some initial approximation to the critical numbers is established by plotting f [x ] ).

so taking  (5.1841) = 5.1841 gives the correct maximum attained at x = 0.902652.

Absolute Extrema on  [a,b]

Continuous function on a closed bounded interval attains its absolute min and absolute max, quite often at the endpoints of the interval  [a,b].  Suspects for absolute min or max are among critical numbers augmented by points a, b.

Newton's Method

To solve  f [x] = 0 numerically (which is always the case for any polynomial of degree higher than 5, except
in the special cases) the Newton's algorithm is as follows : identify the starting point near the solution
and use   = .  Then lim = x  is the root, i.e., f [x]=0.  The initial guess should be found
  from the graph of  f [x].

To find all roots of   = 0,  graph first over large interval [-a, a].

As seen from the graph (zoom if needed) there are 2 real roots, namely around -3 and around 10.  
To find the positive solution take the initial guess as 9:

To streamline computations define a new function g[x] which evaluates numerically the first 25 digits
of approximation sequence:

Then

To compute  n iterations starting at x0 one uses the NestList[g, x0, n] as follows:

Compare the above results to built-in functions such as NRoots[expr[x] = = 0, x], NSolve[ expr[x] = = 0], FindRoot[expr[x] = = 0,{x, }],
where is the initial approximation chosen in the vicinity of the exact solution.

As one can see the 4-th iteration in  Newton's scheme agrees up to the first 7 digits with the exact solution,
while the 6-th iteration provides the exact first 21 digits!  Numerical analysts, "Go to dust!"