math experts... functions?

Question

find two different paris of functions f and g such that h(x) = f(g(x))

h(x) = c, where c is a constant

Answer 1

One pair could be f(x)=c and g(x)=x; then h(x)=f(g(x))=f(x)=c. One more pair could be f(x)=3c-x and g(x)=2c, then
h(x)=f(g(x))=f(2c)=3c-2c=c

Answer 2

The simplest way to create these functions is to make f or g (or both) constant functions like h, where all inputs produce the same output.

For example:

g(x) = x^2 - 5x + 67/pi; f(x) = c.
(f "doesn't care" what its input is.)

OR:
g(x) = 982.3; f(x) = c.
(Both are constant functions.)

OR:
g(x) = c+42; f(x) = x-42.
(g "doesn't care" what its input is; f is set up to adjust g's constant output into c.)


f and g MAY both be nonconstant functions, but f must equal c for all possible values of g(x).

For example -- picking some more unusual functions:

g(x) = { 2 if x is an even integer; 3 if x is an odd integer; 4 otherwise.
f(x) = c + trunc(x/10)

(trunc(x) means the integer part of x, discarding the fractional part. trunc(0.4)=0; trunc(234.56)=234; trunc(-13.01)=-13. IN GENERAL g has more than one output value, but g(2), g(3), and g(4) are all c+0.)


Finally, teachers have been known to check the web! :-)
Don't copy anything from here verbatim; make up your own pairs of functions.

Answer 3

At no aspect do those 2 graphs intersect. the first one sounds like an time-honored y=x graph, yet its shitfed down 3, or maybe as x passes via the x axis, instead of persevering with down it bounces lower back up, as if it were now y=-x-3. the second one graph is merely y=x+a million. They in no way intersect, so in no case does |x-3|=x+a million