How is 2|x|3 -1 built?
Answer: g(f(h(x)))
How is |2x3-1|
built?
Answer: h(g(f(x)))
and
.
Find the domain for f ( g (x))
and again for g ( f (x)).
Domain for f(x):
x
¹
-2
Domain for g(x):
4
- x2 ³ 0
(2 - x)(2 + x) ³ 0
-
+
-
Squiggly Line: -------|----------------|------
-2
2
Answer: -2 £
x
£
2
f ( g (x)):

x--------------------- Ö(4
- x2) -------------------- 1/(Ö(4
- x2) + 2)
Ö(4 - x2) ¹
-2
It never is; square roots are positive values.
Domain: -2 £ x £ 2 because these are the only values you can put into the first function.
g ( f (x):

x---------------------
1/(x + 2) -------------------- Ö(4
- [1/(x + 2)] 2)
-2 £ 1/(x + 2) £
2
1/(x + 2) ³ -2 and
1/(x + 2) £
2
1/(x + 2) + 2 ³ 0
1/(x + 2) - 2 £
0
1 + 2(x + 2) ³ 0
1 - 2(x + 2) £
0
x +
2
x + 2
1 + 2x + 4 ³ 0
1 - 2x - 4 £
0
x +
2
x + 2
2x + 5 ³ 0
- 2x - 3 £
0
x +
2
x + 2
+ -
+
- + -
Squiggly Line:
---|------|-----
----|-----|-----
-5/2
-2
-2 -3/2
x £ -5/2, x > -2
and -2 < x £
-3/2
Domain: -2 < x £
-3/2