H3 #3 a & b

Show that the function g(x) = (3x^2+3)^(1/4) maps the interval [1,2] into [1,2]. In other words, the range of g(x) for 1 =< x =< 2 is contained in [1,2].

Here it seems that we are dealing with simply mapping the two endpoints into the equation:

p0	p1
1	2
(3*(p0)^2+3)^(1/4)	(3*(p1)^2+3)^(1/4)

b. Using the fact that 1 =< x =< 2 show that |g'(x)| =< k for x E [1,2], where k is a constant strictly less than one.


---I don't know exactly but I am playing with the idea that since x is within [1,2] and g(x) seems to map [1,2] to approx [1.56, 1.97] it just makes sense that the derivative will never pass 1...