Find the composite function.

I seem to be struggling with part B of this assignment.

I plot the function into every value of xx, yet it does not help me to solve it at all! Is there a different way to do this? Is the substitution of xx by f(x)f(x) necessary, or does a less time consuming way exist(perhaps doing something to the previously found domain and range of ff)?

Appreciate the help.



1 Answer


Hint (a): to find the range of ff, let y=f(x)=2−x+1×2+2x+2y = f(x) = 2 – \frac{x+1}{x^2+2x+2}. Then:

y(x2+2x+2)=2(x2+2x+2)−(x+1)y(x^2+2x+2)=2(x^2+2x+2) – (x+1)

(y−2)x2+(2y−3)x+2y−3=0(y – 2)x^2 + (2 y – 3) x + 2y -3 = 0

The range of ff is the range of values yy for which the above has real roots.

Δ=(2y−3)2−4(y−2)(2y−3)=−(2y−3)(2y−5)≥0\Delta = (2y-3)^2-4(y-2)(2y-3) = -(2y-3)(2y-5) \;\;\ge\;0

Hint (b): once you determined at step (a) that the range of ff is f(R)=[32,52]f(\mathbb{R})=[\frac{3}{2},\frac{5}{2}], the range of f∘ff \circ f will be f(f(R))=f([32,52])f(f(\mathbb{R}))=f([\frac{3}{2},\frac{5}{2}]). It helps to notice that ff is increasing for x≥0x \ge 0, which becomes more apparent if you write it as:

f(x)=2−1(x+1)+1(x+1)f(x) = 2 – \frac{1}{(x+1)+\cfrac{1}{(x+1)}}