Two calculator exam problems

[CR Haeger]

(05-27-2014 01:47 PM)Peter Murphy Wrote:  I would greatly appreciate seeing solutions to the following problem:

A. Let f(x) = 1 +x(2 - x(3 - x(4 - x(1 + x)))). Find the solution to
f(f(x)) = 1/2 that is nearest to x = 0.


And, especially if they would provide further insight, solutions to this:

B. Let f(x) = sin (1/x) and define g(x) = f(f(f(f(x)))). Find the maximum value of g(x) for x in the range 102 to 103.2.

Many thanks in advance.

Using HP Prime

A. Defining f(x):=1 +x(2 - x(3 - x(4 - x(1 + x)))), then g(x):=f(f(x)) then fsolve(g(x)=1/2, x, -1..1) yields five roots between -1 to +1 with +0.21399.... closest to zero. Defining Function.F1:=g(x)-1/2 lets you view this interesting curve.

B. Similarly. define f(x):=sin(1/x) then g(x):=f(f(f(f(x))). Defining Function.F2:=g(x) lets you graphically explore g(x) in the range 102-103.2 (assumed degrees?). Although 102 appears highest, 103.2 is max at 0.9459... - there seems to be a local max of ~1.00 at ~103.25 degrees..

Best,
Carl