composite function with conditional IF I've been wrapping my head around my Computer and Logic Essentials class, I can do most composite functions, however there is one question that I'm confused with.
It has an if statement inside it:
f: R → R, f(x) = 4x – 3
g: R → R, g(x) = x2 + 1
h: R → R, h(x) = 0 (if x ≥ 0), h(x) = 0 (if x < 0)
determine:
(a) f º f - done, confirmed correct
(b) f º g - done, confirmed correct
(c) h º f - confused
(d) h º g - confused

The first two were easy because they don't have that conditional. One thing i'm confused about is they both say 0 (if condition), on both accounts. Shouldn't they be a different value, otherwise the function will always return a 0?

Secondly, the answer is supposedly:
(c) (h º f) (x) = 1 if x ≥ ¾, (h º f) (x) = 0 if x < ¾
(d) (h º g) (x) = 1

for c, it states 1 if... 0 if... So i'll assume it's a typo in the question (is this a valid assumption?) If so, then h now becomes: h: R → R, h(x) = 1 (if x ≥ 0), h(x) = 0 (if x < 0)


Alright. Now when it comes to the answer, I have no idea how on earth they came up with x is ≥ ¾
What i've come up with so far is:
h o f = h(f(x)) = h(4x-3) 
= 1 (if(4x-3) >= 0, 0 (if(4x-3) < 0)

this seems like the rational first step to the problem, since I should be replacing x with the function as no value is given for x.


Can someone please explain how they came up with the 3/4 figure? And how that was used to solve the question (c)? Thanks.

You already know everything. Yes, it seems a typo.
$$4x-3\ge 0 \iff 4x\ge 3 \iff x\ge 3/4\,.$$
So,  for $h\circ f=\ x\mapsto h(f(x))$, we have to calculate $h(4x-3)$. 
By def. of $h$, this is $1$ if $4x-3\ge 0$, i.e. if $x\ge 3/4$.
For part (d), observe that $x^2+1\ge 1\ge 0$ for all real $x$.
